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# How to manipulate "12 = x - (square root of x)" into "x="

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Hey, everybody. 9 is also a solution to the original problem. Who said there was only one solution? Even though in common parlance, we say something like "3 is

posted by mhum at 10:41 PM on February 23, 2011 [4 favorites]

Gah, you're right, mhum. (the curse of the negative sign forever hangs over me.)

But x = 9 would only be a solution to the original problem if there was a ± sign in front of the sqrt(x)...

posted by btfreek at 11:03 PM on February 23, 2011

No, it isn't. The square root function is always nonnegative. Seriously. Graph it. It is never equal to -3 (as long as we only consider real numbers). Do not chime in with blatantly false comments like this.

A totally valid solution to

What you wrote is not true. How can a single quantity be equal to two different things, which indeed they are not themselves equal? Misunderstandings like this is why people have problems with algebra. If you just write sqrt(a^2), that is not equal to "+a" or "-a". It is equal to the absolute value of a.

Anyway...

As mentioned above, OP, when you square both sides of the equation, you may introduce an extraneous solution. That is why when you get 16 and 9 as your candidates for solutions, you must go back and check to see what works.

When you do algebra like Blue Jello Elf and Hactare, all you're doing is saying the following:

OK, if I have a value of x such that

12 = x - sqrt(x),

then that same value of x must satisfy

12 - x = -sqrt(x).

Indeed, that same value of x must satisfy this equation:

(12-x)^2 = (-sqrt(x))^2.

Indeed, which must satisfy:

144 - 24x + x^2 = x.

Indeed, that same value of x must satisfy

144 -25x + x^2 = 0.

And that same value of x must satisfy

(x-16)(x-9) = 0,

Which tells me that x = 16, or x = 9, or both. Now I have to go check which of those actually works in the original equation I was given.

posted by King Bee at 4:10 AM on February 24, 2011 [4 favorites]

Though maybe they're teaching math differently these days-- back when I solved a square root, it was standard to write sqrt(x) = +/- y. I haven't seen any evidence that this sort of notation has changed.

posted by deanc at 5:57 AM on February 24, 2011

It is true. The radical function has a non-negative range. It results in the principal root. 9 is not a solution to this problem.

posted by monkeymadness at 7:33 AM on February 24, 2011 [1 favorite]

This is one of them. I'd bet dollars to doughnuts that any algebra textbook on the planet that is not written by a hack agrees with me that 16 is the only solution to the given equation. Whatever you think you may have "learned" in the past is just plain wrong.

You want an equation where 9 is the only solution, and square roots are involved? 6 = x - sqrt(x) is perfectly fine (or maybe I don't understand what you're asking for).

You are not remembering correctly about what you learned about "square root means plus or minus". You must

So, if all you know about two real numbers is that their squares are equal, then you cannot claim indeed that the two real numbers I'm thinking of are equal. This is what you are thinking of. You are not thinking "oh yeah, I was totally taught that whenever I see sqrt(x), they are referring to either the positive OR the negative square root of x". That didn't happen. Or, if it did, find out who taught you that and slap them in the face.

There is no difference between convention and notation here. If you see "sqrt(x)", that means principal (positive) square root of the number x.

posted by King Bee at 10:18 AM on February 24, 2011 [1 favorite]

Oh believe me, I sympathize. However I don't agree it's bad form, dirty play or anything like that. I think that when discussing a matter that can be solved in proper technical debate, one should take a good faith assumption on all participants. Particularly when some of them, as seems the case here, are learning the concepts involved. So, if I may be so bold as to make suggestions and hope not to get under your collar again, consider those answers and their authors (me included when I was wrong up there) not as posting blatantly false questions with a poisoning or bragging intention, but as good faith mistakes that can be corrected while trying to keep one's calm and all that. Not easy, I know (my teaching experience on Comp Sci has granted me much less hair and grayer than I should have at my age, heh) but always worth trying?

(Cultural aside: seems to be much less the case these days, I'd need to browse what they are giving the kids on precalculus here right now, but we were actually encouraged in many courses to use the square root sign and the 1/2 exponent as literally the same thing in different notation, and I know this was still teh case many years later, judging from when I began to teach students in uni and I had to work around that notion. So I idly wondered up there if something like that might not be the case with some of the answers. Our math teachers have always had some strange off-the-regular-track ideas here, not sure what is the actual origin of the unorthodox teaching plans that the governments have pushed around us since I was a kid.)

posted by Iosephus at 12:58 PM on February 24, 2011

But, this was sort of my point -- that in certain contexts sqrt(x) is not considered a function. This appears to be a rather minority view (to say the least) and the contexts where this would be the case appear to have mostly vanished. One of these contexts appears to be in (old) IMO problems. Consider this problem from 1959 which explicitly excludes negative roots in the problem statement. Or, consider this problem from 1963, which handles a case of negative roots in the solution.

I can't help but wonder if there's a regional component here (e.g.: American vs. non-American practice) or if it's simply the march of time.

posted by mhum at 1:33 PM on February 24, 2011

Note that they say "

posted by mhum at 2:24 PM on February 24, 2011

I'm not sure I was clear on this point. I was implying that since for every solution x<0>0, we could wlog assume x>0. We couldn't change the sign of only one root because it wouldn't guarantee that we would preserve a solution.

Of course, all of this is moot. Upon closer inspection, I would have to agree that they don't seem to be considering the possibility of negative roots. The wording in the exposition is, in my opinion, really not great. I would have used "since" instead of "if" and "however".

It does appear that the only time people raise the possibility of a surd representing a negative root is only to rule it out (see also Question 2 in the 1990 USAMO). So, aside from a century-old text (classic though it may be), there really is no significant support for my previous position.

posted by mhum at 7:33 PM on February 24, 2011

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# How to manipulate "12 = x - (square root of x)" into "x="

February 23, 2011 8:56 PM Subscribe

The solution to 12 = x - (square root of x) is 16, but how do I manipulate the equation to show that?

I've been tiring myself out trying to manipulate this equation down to an "x =" form for too long, so I thought I'd try opening this up to the hive mind.

My first thought was that you could simply square both sides, but that leaves you with 144 = (x-sqrtx)(x-sqrtx) x^2 - 2*x^1.5 + x (not so helpful.) This is a frustrating problem for me because it's so obvious that it equals 16, but I can't make the math say so.

(I apologize in advance if the idiocy of this question offends all of the math majors and Ph.D. candidates that hang around here. I haven't taken a math class in a long time.)

I've been tiring myself out trying to manipulate this equation down to an "x =" form for too long, so I thought I'd try opening this up to the hive mind.

My first thought was that you could simply square both sides, but that leaves you with 144 = (x-sqrtx)(x-sqrtx) x^2 - 2*x^1.5 + x (not so helpful.) This is a frustrating problem for me because it's so obvious that it equals 16, but I can't make the math say so.

(I apologize in advance if the idiocy of this question offends all of the math majors and Ph.D. candidates that hang around here. I haven't taken a math class in a long time.)

Subtract x from both sides.

12-x = sqrt{x}

square both sides

(12-x)^2=x

x^2-24x+144=x

x^2-25x+144=0

(x-16)(x-9)=0

9 does not work, so you're left with x=16

posted by Hactar at 9:02 PM on February 23, 2011 [1 favorite]

12-x = sqrt{x}

square both sides

(12-x)^2=x

x^2-24x+144=x

x^2-25x+144=0

(x-16)(x-9)=0

9 does not work, so you're left with x=16

posted by Hactar at 9:02 PM on February 23, 2011 [1 favorite]

Or, you could factor out a sqrt(x) from the right hand side, leaving you with 12 = (sqrt(x) - 1)*(sqrt (x)). Now substitute, say, "u" for sqrt (x),

12 = (u-1)*u

u = 4

sqrt(x) = 4

x = 16

posted by btfreek at 9:04 PM on February 23, 2011

12 = (u-1)*u

u = 4

sqrt(x) = 4

x = 16

posted by btfreek at 9:04 PM on February 23, 2011

That should be 12-x=-sqrt{x}, but you're squaring, which makes the missed negative irrelevant.

posted by Hactar at 9:04 PM on February 23, 2011

posted by Hactar at 9:04 PM on February 23, 2011

12 = x - sqrt(x)

12 - x = x - sqrt(x) - x

12 - x = -sqrt(x)

(12-x)^2 = (-sqrt(x))^2

144 - 24x + x^2 = x

144 - 25x + x^2 = 0

Then you can apply the quadratic formula, which states that ax^2 + bx + c = 0 has the following solutions:

x = (-b +/- sqrt( b^2 - 4ac)) / 2a

= ( 25 +/- sqrt( 625 - 576) ) / 2

= ( 25 +/- sqrt( 49 ) ) / 2

= ( 25 +/- 7 ) / 2

= 18/2 or 32 / 2

= 9 or 16

x=9 doesn't satisfy the original equation, because on the line where we squared both sides, we introduced some ambiguity: we "lost" the negative sign, and x=9 is actually the solution to 12=x+sqrt(x).

posted by Blue Jello Elf at 9:10 PM on February 23, 2011

12 - x = x - sqrt(x) - x

12 - x = -sqrt(x)

(12-x)^2 = (-sqrt(x))^2

*Note: this step introduces issues that I explain later*144 - 24x + x^2 = x

144 - 25x + x^2 = 0

Then you can apply the quadratic formula, which states that ax^2 + bx + c = 0 has the following solutions:

x = (-b +/- sqrt( b^2 - 4ac)) / 2a

= ( 25 +/- sqrt( 625 - 576) ) / 2

= ( 25 +/- sqrt( 49 ) ) / 2

= ( 25 +/- 7 ) / 2

= 18/2 or 32 / 2

= 9 or 16

x=9 doesn't satisfy the original equation, because on the line where we squared both sides, we introduced some ambiguity: we "lost" the negative sign, and x=9 is actually the solution to 12=x+sqrt(x).

posted by Blue Jello Elf at 9:10 PM on February 23, 2011

Both of these solutions are making leaps that I don't understand

@Hactar: shouldn't (x-16)(x-9)=0 give you x = 16 AND x = 9 as the solution?

@btfreek: how do you simplify 12 = u(u-1) into u=4?

posted by Wanderboy at 9:12 PM on February 23, 2011

@Hactar: shouldn't (x-16)(x-9)=0 give you x = 16 AND x = 9 as the solution?

@btfreek: how do you simplify 12 = u(u-1) into u=4?

posted by Wanderboy at 9:12 PM on February 23, 2011

Once you get it into standard form x^2-25x+144=0 form (as Hactar showed) there are lots of ways to solve. You could also use the Quadratic equation.

posted by Diplodocus at 9:12 PM on February 23, 2011

posted by Diplodocus at 9:12 PM on February 23, 2011

To get from 12=u(u-1) to u = 4, just rewrite as u^2 - u - 12 = 0 and solve using the quadratic formula. Or in this case ask yourself which two consecutive numbers can be multiplied to get 12 - that's 3 and 4.

posted by madcaptenor at 9:16 PM on February 23, 2011

posted by madcaptenor at 9:16 PM on February 23, 2011

12 = u(u-1)

12 = u^2 - u

u^2 - u - 12 = 0

Now factor:

(u +3)(u - 4) = 0

u = -3 or +4

You can't take the square root of a negative number, so u must be 4.

posted by btfreek at 9:17 PM on February 23, 2011

12 = u^2 - u

u^2 - u - 12 = 0

Now factor:

(u +3)(u - 4) = 0

u = -3 or +4

You can't take the square root of a negative number, so u must be 4.

posted by btfreek at 9:17 PM on February 23, 2011

Yes, brfreek spelled out what I was trying to get at (and I'm not sure it's not called completing the square...that's something else...been a while for me, too... it's just factoring.)

posted by Green Eyed Monster at 9:22 PM on February 23, 2011

posted by Green Eyed Monster at 9:22 PM on February 23, 2011

When you square the equation, you introduce an extraneous solution. If you plug in 9, it works only if you say the square root of 9 is -3. (-3)^2 is 9, but the generally accepted square root of 9 is 3, so we cannot use x=9 as a solution.

When I squared the equation in my previous comment, I got rid of any thought of a positive or negative root, which is where the extra apparent solution came from.

posted by Hactar at 9:27 PM on February 23, 2011 [2 favorites]

When I squared the equation in my previous comment, I got rid of any thought of a positive or negative root, which is where the extra apparent solution came from.

posted by Hactar at 9:27 PM on February 23, 2011 [2 favorites]

Oh, and if you need the steps for factoring, it's probably easiest to watch an explanation instead of reading it.

(Completing the square is factoring when it doesn't all slot nicely in place, iirc. All I remember is that I hated doing it and would rather just take the quadratic formula any day)

posted by btfreek at 9:29 PM on February 23, 2011

(Completing the square is factoring when it doesn't all slot nicely in place, iirc. All I remember is that I hated doing it and would rather just take the quadratic formula any day)

posted by btfreek at 9:29 PM on February 23, 2011

(yeah, completing the square would be if it was x^2+10x and you had to +25 to each side to be able to factor it...and I would go out of my way to avoid the quadratic equation.)

posted by Green Eyed Monster at 9:47 PM on February 23, 2011

posted by Green Eyed Monster at 9:47 PM on February 23, 2011

Here's an approach to factoring it using the FOIL method, which is a helpful tool to remember if you are just getting back into math.

Move the 12 to the right side of the equation:

0 = x - sqrt (x) - 12

Which looks like a quadratic, right? Now think about "first-outside-inside-last."

For the "first" step, what can you multiply with itself to get 'x'? The square root of x.

So you could rewrite this equation as:

0 = (sqrt(x))^2 - sqrt(x) - 12

This takes the form ax^2 + bx + c, which you are probably familiar with.

Now for the "inside" and "outside" steps, what two numbers can each be multiplied with sqrt (x), then those products added to each other, to get '-sqrt (x)'?

Any two numbers where one is negative, and greater than the other in absolute value by one, right? (1,-2; 2,-3; 3,-4; etc)

Now for the "last" step, which pair of those numbers can also be multiplied together to get -12?

Only one pair: 3 and -4.

You can rewrite your equation now as:

0 = (sqrt (x) + 3) (sqrt (x) - 4)

The 'zeroes' of that equation are where sqrt (x) = 4 at x = 16, and where sqrt (x) = -3, which has no real solutions.

So your only real solution is x=16.

posted by edguardo at 9:53 PM on February 23, 2011

Move the 12 to the right side of the equation:

0 = x - sqrt (x) - 12

Which looks like a quadratic, right? Now think about "first-outside-inside-last."

For the "first" step, what can you multiply with itself to get 'x'? The square root of x.

So you could rewrite this equation as:

0 = (sqrt(x))^2 - sqrt(x) - 12

This takes the form ax^2 + bx + c, which you are probably familiar with.

Now for the "inside" and "outside" steps, what two numbers can each be multiplied with sqrt (x), then those products added to each other, to get '-sqrt (x)'?

Any two numbers where one is negative, and greater than the other in absolute value by one, right? (1,-2; 2,-3; 3,-4; etc)

Now for the "last" step, which pair of those numbers can also be multiplied together to get -12?

Only one pair: 3 and -4.

You can rewrite your equation now as:

0 = (sqrt (x) + 3) (sqrt (x) - 4)

The 'zeroes' of that equation are where sqrt (x) = 4 at x = 16, and where sqrt (x) = -3, which has no real solutions.

So your only real solution is x=16.

posted by edguardo at 9:53 PM on February 23, 2011

*9 does not work*

*x=9 doesn't satisfy the original equation*

*shouldn't (x-16)(x-9)=0 give you x = 16 AND x = 9 as the solution*

*If you plug in 9, it works only if you say the square root of 9 is -3*

Hey, everybody. 9 is also a solution to the original problem. Who said there was only one solution? Even though in common parlance, we say something like "3 is

**the**square root of 9", it would be slightly more accurate to say that "3 is

**a**square root of 9"; -3 is, of course, the other square root of 9. Every positive number has two square roots, a positive one and a negative one. The statement of the equation does not specify that the square root in question is positive.

posted by mhum at 10:41 PM on February 23, 2011 [4 favorites]

btfreek:

The factor (u-3) is not implying taking the square root of a negative number. Instead, it's implying that any x where sqrt(x) = -3 is a solution to the equation. It's not saying that any x where sqrt(-3) = x is a solution. A totally valid solution to sqrt(x) = -3 is x = 9.

posted by mhum at 10:47 PM on February 23, 2011

*You can't take the square root of a negative number, so u must be 4.*The factor (u-3) is not implying taking the square root of a negative number. Instead, it's implying that any x where sqrt(x) = -3 is a solution to the equation. It's not saying that any x where sqrt(-3) = x is a solution. A totally valid solution to sqrt(x) = -3 is x = 9.

posted by mhum at 10:47 PM on February 23, 2011

*A totally valid solution to sqrt(x) = -3 is x = 9.*

Gah, you're right, mhum. (the curse of the negative sign forever hangs over me.)

But x = 9 would only be a solution to the original problem if there was a ± sign in front of the sqrt(x)...

posted by btfreek at 11:03 PM on February 23, 2011

Whenever you get a possible solution you have to check it against the original statement to see if it actually is a solution. Sqrt, the symbol, results in a non-negative (the range doesn't include negatives) so 9 is not a solution.

posted by monkeymadness at 3:46 AM on February 24, 2011

posted by monkeymadness at 3:46 AM on February 24, 2011

Use a change of variables and then the quadratic equation (or completing the square).

Original equation: x - sqrt(x) = 12

Rearrange: x - sqrt(x) - 12 = 0

Change of variable: let y=sqrt(x), so that y^2 - y = 12

The equation becomes: y^2 - y - 12 = 0

Applying the quadratic equation gives y=(1 +/- 7)/2

ie. y = 4 or y = -3

which implies (by reversing the change of variables, by squaring y) that x = 16 or x = 9.

Check both answers are valid:

x=16: 16 - sqrt(16) = 16 - 4 = 12 -- okay (taking the positive square root)

x=9: 9 - sqrt(9) = 9 + 3 = 12 -- okay (taking the negative square root)

("

posted by manyon at 3:56 AM on February 24, 2011 [1 favorite]

Original equation: x - sqrt(x) = 12

Rearrange: x - sqrt(x) - 12 = 0

Change of variable: let y=sqrt(x), so that y^2 - y = 12

The equation becomes: y^2 - y - 12 = 0

Applying the quadratic equation gives y=(1 +/- 7)/2

ie. y = 4 or y = -3

which implies (by reversing the change of variables, by squaring y) that x = 16 or x = 9.

Check both answers are valid:

x=16: 16 - sqrt(16) = 16 - 4 = 12 -- okay (taking the positive square root)

x=9: 9 - sqrt(9) = 9 + 3 = 12 -- okay (taking the negative square root)

("

*Sqrt, the symbol, results in a non-negative*": not true; sqrt(a^2) = +a**-a)***or*posted by manyon at 3:56 AM on February 24, 2011 [1 favorite]

*A totally valid solution to sqrt(x) = -3 is x = 9.*

No, it isn't. The square root function is always nonnegative. Seriously. Graph it. It is never equal to -3 (as long as we only consider real numbers). Do not chime in with blatantly false comments like this.

A totally valid solution to

*x^2 = 9*is x=3, and so is x=-3. That is what you are thinking of.

*("Sqrt, the symbol, results in a non-negative": not true; sqrt(a^2) = +a or -a)*

What you wrote is not true. How can a single quantity be equal to two different things, which indeed they are not themselves equal? Misunderstandings like this is why people have problems with algebra. If you just write sqrt(a^2), that is not equal to "+a" or "-a". It is equal to the absolute value of a.

Anyway...

As mentioned above, OP, when you square both sides of the equation, you may introduce an extraneous solution. That is why when you get 16 and 9 as your candidates for solutions, you must go back and check to see what works.

When you do algebra like Blue Jello Elf and Hactare, all you're doing is saying the following:

OK, if I have a value of x such that

12 = x - sqrt(x),

then that same value of x must satisfy

12 - x = -sqrt(x).

Indeed, that same value of x must satisfy this equation:

(12-x)^2 = (-sqrt(x))^2.

Indeed, which must satisfy:

144 - 24x + x^2 = x.

Indeed, that same value of x must satisfy

144 -25x + x^2 = 0.

And that same value of x must satisfy

(x-16)(x-9) = 0,

Which tells me that x = 16, or x = 9, or both. Now I have to go check which of those actually works in the original equation I was given.

posted by King Bee at 4:10 AM on February 24, 2011 [4 favorites]

mhum:

You're right. It clearly specifies that you should be taking the negative square root. It totally says it

OP, trust me. I am correct, people who are saying that 9 is also a solution to the original equation are incorrect.

posted by King Bee at 4:16 AM on February 24, 2011 [1 favorite]

*The statement of the equation does not specify that the square root in question is positive.*You're right. It clearly specifies that you should be taking the negative square root. It totally says it

*right there*.OP, trust me. I am correct, people who are saying that 9 is also a solution to the original equation are incorrect.

posted by King Bee at 4:16 AM on February 24, 2011 [1 favorite]

The radical sign indicates what is called the principal square root (the positive one), not any square root, so 9 won't fly. Squaring both sides of the equation gave an equation that had an extra root ( since the same equation would result from squaring with a minus sign before the radical sign.)

posted by Obscure Reference at 4:30 AM on February 24, 2011 [2 favorites]

posted by Obscure Reference at 4:30 AM on February 24, 2011 [2 favorites]

Like most of us who have mastered Boolean algebra, he is aware that "or" represents one thing out of two, not two different things.("Sqrt, the symbol, results in a non-negative": not true; sqrt(a^2) = +a or -a)

What you wrote is not true. How can a single quantity be equal to two different things, which indeed they are not themselves equal?

Though maybe they're teaching math differently these days-- back when I solved a square root, it was standard to write sqrt(x) = +/- y. I haven't seen any evidence that this sort of notation has changed.

posted by deanc at 5:57 AM on February 24, 2011

"Every non-negative real number x has a unique non-negative square root, called the principal square root, denoted by a radical sign as \scriptstyle \sqrt{x}."

http://en.wikipedia.org/wiki/Square_root

posted by 23skidoo at 6:37 AM on February 24, 2011 [1 favorite]

http://en.wikipedia.org/wiki/Square_root

posted by 23skidoo at 6:37 AM on February 24, 2011 [1 favorite]

deanc:

Indeed. However, if we are considering an equation like sqrt(x^2) = -1, there is no solution (since we must respect the order of operations). What value of x would you choose, such that when you square it and take its principal square root, you would get a negative number? You won't find one.

sqrt(x^2) = |x|, not "plus or minus x".

This is the difference between me writing sqrt(x^2) and "taking square roots of both sides of an equation". Like most of us who have mastered elementary algebra, we are aware that there is a difference.

posted by King Bee at 7:02 AM on February 24, 2011

*Like most of us who have mastered Boolean algebra, he is aware that "or" represents one thing out of two, not two different things.*Indeed. However, if we are considering an equation like sqrt(x^2) = -1, there is no solution (since we must respect the order of operations). What value of x would you choose, such that when you square it and take its principal square root, you would get a negative number? You won't find one.

sqrt(x^2) = |x|, not "plus or minus x".

This is the difference between me writing sqrt(x^2) and "taking square roots of both sides of an equation". Like most of us who have mastered elementary algebra, we are aware that there is a difference.

posted by King Bee at 7:02 AM on February 24, 2011

*not true; sqrt(a^2) = +a or -a*

It is true. The radical function has a non-negative range. It results in the principal root. 9 is not a solution to this problem.

posted by monkeymadness at 7:33 AM on February 24, 2011 [1 favorite]

Hmm. It appears that there is a difference in convention and notation -- namely, whether to treat "square root" as the function representing the principal square root (as King Bee is strongly advocating for), or as the one-to-many map that associates a number with all of its roots (as I advised earlier). There are, of course, many contexts in which there is no ambiguity, e.g.: if I apply Pythagoras to deduce that the length of a hypotenuse is sqrt(a^2 + b^2), I'm pretty sure I'm not talking about negative lengths. In my interpretation of the problem in an algebraic context, I had inferred the one-to-many form of the square root.

To those who contend that 16 is the one and only solution to the equation, I would ask how we would re-write the equation so that 9 was the one and only solution in this notation.

posted by mhum at 8:35 AM on February 24, 2011

To those who contend that 16 is the one and only solution to the equation, I would ask how we would re-write the equation so that 9 was the one and only solution in this notation.

posted by mhum at 8:35 AM on February 24, 2011

mhum: I don't want to weigh in on the whole "what does sqrt(x) really mean" issue. But I'd suggest x + sqrt(x) = 12 as a rewriting where 9 is the only solution.

posted by madcaptenor at 8:38 AM on February 24, 2011 [1 favorite]

posted by madcaptenor at 8:38 AM on February 24, 2011 [1 favorite]

madcaptenor, if we use the non-principle roots, x+sqrt{x} = 12 can be solved with 16 and -4.

The main reason not to use anything other than principle roots (in my opinion, at least) is what happens when you start using complex numbers. 1^{1/3} (The third root of 1) has one solution in the real numbers (1) two more complex solutions ( (-sqrt{2}+i)/(3) and (-sqrt{2}-i)/(3)) (this is assuming that I did the math right in my head). By including solutions that are not the principle root in this sort of problem you can introduce a whole host of other issues. If you want to go traipsing into the complex plane, you can, but that does open a whole other can of worms.

posted by Hactar at 9:09 AM on February 24, 2011

The main reason not to use anything other than principle roots (in my opinion, at least) is what happens when you start using complex numbers. 1^{1/3} (The third root of 1) has one solution in the real numbers (1) two more complex solutions ( (-sqrt{2}+i)/(3) and (-sqrt{2}-i)/(3)) (this is assuming that I did the math right in my head). By including solutions that are not the principle root in this sort of problem you can introduce a whole host of other issues. If you want to go traipsing into the complex plane, you can, but that does open a whole other can of worms.

posted by Hactar at 9:09 AM on February 24, 2011

*There are, of course, many contexts in which there is no ambiguity*

This is one of them. I'd bet dollars to doughnuts that any algebra textbook on the planet that is not written by a hack agrees with me that 16 is the only solution to the given equation. Whatever you think you may have "learned" in the past is just plain wrong.

You want an equation where 9 is the only solution, and square roots are involved? 6 = x - sqrt(x) is perfectly fine (or maybe I don't understand what you're asking for).

You are not remembering correctly about what you learned about "square root means plus or minus". You must

*consider*the positive and negative square roots when you are trying to "solve for x" in a given equation and you want to "take the square root of both sides". The reason why you must do this is because the squaring function is not one-to-one, and therefore has no inverse (or "undoing" operation) like the "add six" or "divide by 50" operations do.

So, if all you know about two real numbers is that their squares are equal, then you cannot claim indeed that the two real numbers I'm thinking of are equal. This is what you are thinking of. You are not thinking "oh yeah, I was totally taught that whenever I see sqrt(x), they are referring to either the positive OR the negative square root of x". That didn't happen. Or, if it did, find out who taught you that and slap them in the face.

There is no difference between convention and notation here. If you see "sqrt(x)", that means principal (positive) square root of the number x.

posted by King Bee at 10:18 AM on February 24, 2011 [1 favorite]

Some technical nitpicking that the OP can safely ignore, I guess:

- yes the problem changes if you take sqrt as the "principal root function" or as a typographic convetion for raising to the 1/2 power.

- yes the problem also changes if you allow complex roots besides real ones. The easiest way one avoids getting bogged in that discussion is by preceding the problem statement with something indicating that your polynomials (which is what the original formula is, in a way) are formed over R (reals) or C (complex). Or even more to the point "find the real value of unknown x in the following" versus "find the complex values of unknown x in the following". (Supermathgeeky explanation: why this works for both kind of coefficients is explained by the theory of polynomial rings - hairy details for instance here: http://en.wikipedia.org/wiki/Polynomial_ring)

- "Like most of us who have mastered Boolean algebra, he is aware that "or" represents one thing out of two, not two different things."

This is incorrect or (at best) poorly worded. The standard "or" operator in BA is inclusive, so if I write "2 is even or 7 is bigger than 0", it is a true formula according the semantics of BA. The "one thing out of two" semantics is captured by the usually called "xor" operator, exculsive or.

posted by Iosephus at 11:28 AM on February 24, 2011

- yes the problem changes if you take sqrt as the "principal root function" or as a typographic convetion for raising to the 1/2 power.

- yes the problem also changes if you allow complex roots besides real ones. The easiest way one avoids getting bogged in that discussion is by preceding the problem statement with something indicating that your polynomials (which is what the original formula is, in a way) are formed over R (reals) or C (complex). Or even more to the point "find the real value of unknown x in the following" versus "find the complex values of unknown x in the following". (Supermathgeeky explanation: why this works for both kind of coefficients is explained by the theory of polynomial rings - hairy details for instance here: http://en.wikipedia.org/wiki/Polynomial_ring)

- "Like most of us who have mastered Boolean algebra, he is aware that "or" represents one thing out of two, not two different things."

This is incorrect or (at best) poorly worded. The standard "or" operator in BA is inclusive, so if I write "2 is even or 7 is bigger than 0", it is a true formula according the semantics of BA. The "one thing out of two" semantics is captured by the usually called "xor" operator, exculsive or.

posted by Iosephus at 11:28 AM on February 24, 2011

Yes, Iosephus, the OP should ignore most of what you wrote.

- The square root function symbol is not a "typographic convention" for an exponent of 1/2. It just isn't.

- The problem does not change if we allow x to be complex (instead of pure real). x=16 is still the only solution. Nothing that I did in my solution assumes x is real. All it assumes is that there are no zero divisors in whatever ring I'm working over, so when I have (x-16)(x-9)=0, I can assume that one of the factors must be 0.

- Lastly, come on people. Even Wolfram research agrees with me, and Mathematica is better at everything than all of us combined.

posted by King Bee at 11:56 AM on February 24, 2011

- The square root function symbol is not a "typographic convention" for an exponent of 1/2. It just isn't.

- The problem does not change if we allow x to be complex (instead of pure real). x=16 is still the only solution. Nothing that I did in my solution assumes x is real. All it assumes is that there are no zero divisors in whatever ring I'm working over, so when I have (x-16)(x-9)=0, I can assume that one of the factors must be 0.

- Lastly, come on people. Even Wolfram research agrees with me, and Mathematica is better at everything than all of us combined.

posted by King Bee at 11:56 AM on February 24, 2011

Well, I'll just chalk the first disagreement to different educational backgrounds. I've seen plenty of (excellent) textbooks that went happily along that typographic convention. Perhaps it's a deformation of calculus and algebra in Latin America, I know we have enough of those lurking around. As for the second part, sorry I was confusing, I was speaking in general about "this kind of problem" (see that I never mention the original formula), so let's just say it's my fault for writing too loosely and leave it at that. There's no need to be abrasive and snarky, I know that's getting unusual for the whole site, but I would hope we can get an agreement on that being bad form at least on the math questions. Geez.

posted by Iosephus at 12:12 PM on February 24, 2011

posted by Iosephus at 12:12 PM on February 24, 2011

The point is that the square root symbol isn't a typographic convention for an exponent of 1/2. That sqrt(x) = x^(/12) is true is something, but it's not that one is a shorthand for the other.

I guess I apologize that when you were referring to "the problem", you weren't talking about the one we were talking about. I don't really read anything in my second bullet point to you as necessarily snarky, but whatever.

Snark aside, it's even worse form to post blatantly false things like those that are contained in many of the above "answers" to the OP's original question. That is why I got so hot under the collar.

posted by King Bee at 12:19 PM on February 24, 2011

I guess I apologize that when you were referring to "the problem", you weren't talking about the one we were talking about. I don't really read anything in my second bullet point to you as necessarily snarky, but whatever.

Snark aside, it's even worse form to post blatantly false things like those that are contained in many of the above "answers" to the OP's original question. That is why I got so hot under the collar.

posted by King Bee at 12:19 PM on February 24, 2011

King Bee:

Perhaps. On the other hand, allow me to point to pages 49 and 50 of G. H. Hardy's

Nonetheless, upon further research, I think I may need to concede your point. It appears that the current, common usage of the surd symbol refers primarily -- if not exclusively -- to the function corresponding to the principal root.

posted by mhum at 12:26 PM on February 24, 2011

*I'd bet dollars to doughnuts that any algebra textbook on the planet that is not written by a hack agrees with me that 16 is the only solution to the given equation.*Perhaps. On the other hand, allow me to point to pages 49 and 50 of G. H. Hardy's

*A Course of Pure Mathematics*:It should be noted that there is some ambiguity of notation in such an equation as y = sqrt(x). We have, up to the present, regard (And, you'll find instances (e.g.: page 193, page 443) where he explicitly specifies that an expression containing a surd symbol refers to the positive square root.e.g.) sqrt(2) as denoting thepositivesquare root of 2, and it would be natural to denote by sqrt(x), where x is any positive number, the positive square root of x, in which case y=sqrt(x) would be a one-valued function of x. It is however often more convenient to regard sqrt(x) as standing for the two-valued function whose two values are the positive and negative square roots of x.

Nonetheless, upon further research, I think I may need to concede your point. It appears that the current, common usage of the surd symbol refers primarily -- if not exclusively -- to the function corresponding to the principal root.

posted by mhum at 12:26 PM on February 24, 2011

Alas, keep reading:

In other words, Hardy; what you're talking about ceases to be a function. (Also, luckily I don't consider that book an "algebra textbook", since you generally don't want to bet against Hardy.)

Your concession make me want to concede that when the symbol was introduced

posted by King Bee at 12:50 PM on February 24, 2011

*The reader will observe that, when this course is adopted, the function sqrt(x)***differs fundamentally from the rational functions in two respects**...Secondly, the function, when x has a value for which it is defined, has**generally two values of opposite signs.**In other words, Hardy; what you're talking about ceases to be a function. (Also, luckily I don't consider that book an "algebra textbook", since you generally don't want to bet against Hardy.)

Your concession make me want to concede that when the symbol was introduced

*in 1525*it stood for both positive and negative square root, and that my interpretation is a more "current" one indeed.posted by King Bee at 12:50 PM on February 24, 2011

*Snark aside, it's even worse form to post blatantly false things like those that are contained in many of the above "answers" to the OP's original question. That is why I got so hot under the collar.*

Oh believe me, I sympathize. However I don't agree it's bad form, dirty play or anything like that. I think that when discussing a matter that can be solved in proper technical debate, one should take a good faith assumption on all participants. Particularly when some of them, as seems the case here, are learning the concepts involved. So, if I may be so bold as to make suggestions and hope not to get under your collar again, consider those answers and their authors (me included when I was wrong up there) not as posting blatantly false questions with a poisoning or bragging intention, but as good faith mistakes that can be corrected while trying to keep one's calm and all that. Not easy, I know (my teaching experience on Comp Sci has granted me much less hair and grayer than I should have at my age, heh) but always worth trying?

(Cultural aside: seems to be much less the case these days, I'd need to browse what they are giving the kids on precalculus here right now, but we were actually encouraged in many courses to use the square root sign and the 1/2 exponent as literally the same thing in different notation, and I know this was still teh case many years later, judging from when I began to teach students in uni and I had to work around that notion. So I idly wondered up there if something like that might not be the case with some of the answers. Our math teachers have always had some strange off-the-regular-track ideas here, not sure what is the actual origin of the unorthodox teaching plans that the governments have pushed around us since I was a kid.)

posted by Iosephus at 12:58 PM on February 24, 2011

More often than not, using sqrt(x) and x^(1/2) interchangeably will really not get you into any trouble at all. What I took offense to I think was the "typographical convention" remark. Apologies.

What originally got the vitriol flowing was comments like

and

posted by King Bee at 1:11 PM on February 24, 2011

What originally got the vitriol flowing was comments like

*Hey, everybody. 9 is also a solution to the original problem. Who said there was only one solution?*and

*"Sqrt, the symbol, results in a non-negative": not true; sqrt(a^2) = +a or -a,*

when actually sqrt(a^2) = |a|. As I tried to explain multiple times above, it's the difference in writing "sqrt(x)" and "taking square roots of both sides of an equation".

After that though, pretty much anyone who came in here and tried to convince the OP that x=9 is a perfectly valid solution to the original equation I saw as an enemy combatant. I probably should just relax a little.when actually sqrt(a^2) = |a|. As I tried to explain multiple times above, it's the difference in writing "sqrt(x)" and "taking square roots of both sides of an equation".

After that though, pretty much anyone who came in here and tried to convince the OP that x=9 is a perfectly valid solution to the original equation I saw as an enemy combatant. I probably should just relax a little.

posted by King Bee at 1:11 PM on February 24, 2011

*Secondly, the function, when x has a value for which it is defined, has generally two values of opposite signs.*

But, this was sort of my point -- that in certain contexts sqrt(x) is not considered a function. This appears to be a rather minority view (to say the least) and the contexts where this would be the case appear to have mostly vanished. One of these contexts appears to be in (old) IMO problems. Consider this problem from 1959 which explicitly excludes negative roots in the problem statement. Or, consider this problem from 1963, which handles a case of negative roots in the solution.

I can't help but wonder if there's a regional component here (e.g.: American vs. non-American practice) or if it's simply the march of time.

posted by mhum at 1:33 PM on February 24, 2011

Yeah, in that second problem, they have to throw out the case where

Indeed, that second problem that you mention is dealing with the case where they have the square of a number x equal to some quantity; that means x itself is either the positive square root of that quantity or the negative square root of that quantity. This is the "taking square roots of both sides" thing that I mentioned above. At no point in that solution do I see them acknowledging that the square root symbol means anything but the principal square root.

In fact, they end up doing more like something we did here; they got two

Perhaps this interpretation has a regional component, I don't know anything about that.

posted by King Bee at 2:09 PM on February 24, 2011

*p + 4*is less than*4x^2*because then you would have the square root of a quantity equal to a negative number.Indeed, that second problem that you mention is dealing with the case where they have the square of a number x equal to some quantity; that means x itself is either the positive square root of that quantity or the negative square root of that quantity. This is the "taking square roots of both sides" thing that I mentioned above. At no point in that solution do I see them acknowledging that the square root symbol means anything but the principal square root.

In fact, they end up doing more like something we did here; they got two

*candidates*for solutions, and found out that one leads to a contradiction, so they threw it out (just like we did with our x=9 candidate).Perhaps this interpretation has a regional component, I don't know anything about that.

posted by King Bee at 2:09 PM on February 24, 2011

*At no point in that solution do I see them acknowledging that the square root symbol means anything but the principal square root.*

Note that they say "

**If**we have $p+4 \geq 4x^2$". The part following "However, this is only a solution..." exists to rule out the possibility that $4\sqrt{(x^2-p)(x^2-1)}$ is negative. The contradiction doesn't come from the fact that $4\sqrt{(x^2-p)(x^2-1)}$ is found to be negative; it comes from the fact that if it were negative, it would imply $\sqrt{x^2 - p} \geq x$.

posted by mhum at 2:24 PM on February 24, 2011

It's OK, I can understand unrendered LaTeX just fine. =)

Right, it exists to rule out the possibility that $4\sqrt{(x^2-p)(x^2-1)}$ could possibly be considered to be negative.

The contradiction near the end is that we cannot consider negative values of $p$; otherwise the left hand side of the original equation is

At the very beginning, they also assume nonnegativity of $x$; this is because $x$ is the right hand side of the equation. The left hand side is the sum of two square roots; hence it is nonnegative. This isn't a "case"; it is the only case to consider (since the other one doesn't make sense).

Nowhere in the solution do they consider a negative value for $x$; this is because the square root symbol means principal square root indeed. I'm only bringing this kind of thing up because that second problem you linked to doesn't support your original point, it supports mine.

posted by King Bee at 2:48 PM on February 24, 2011

Right, it exists to rule out the possibility that $4\sqrt{(x^2-p)(x^2-1)}$ could possibly be considered to be negative.

The contradiction near the end is that we cannot consider negative values of $p$; otherwise the left hand side of the original equation is

*necessarily*larger than the right hand side*precisely*because $2\sqrt{x^2-1}$ is nonnegative (for any value of $x$ for which it is defined).At the very beginning, they also assume nonnegativity of $x$; this is because $x$ is the right hand side of the equation. The left hand side is the sum of two square roots; hence it is nonnegative. This isn't a "case"; it is the only case to consider (since the other one doesn't make sense).

Nowhere in the solution do they consider a negative value for $x$; this is because the square root symbol means principal square root indeed. I'm only bringing this kind of thing up because that second problem you linked to doesn't support your original point, it supports mine.

posted by King Bee at 2:48 PM on February 24, 2011

Hmm. I interpreted the exposition slightly differently. I saw the "Assume x>0" part to imply that for any solution with x<0 we have the symmetric solution for x>0 (i.e.: since x only appears as x^2 on the left side, we can change the sign of x and both roots -- simultaneously -- without harm).

I took the "

posted by mhum at 3:13 PM on February 24, 2011

I took the "

**If**p+4>=4x^2" part to imply that the 4*sqrt(...) expression*could*be negative. If it's always the case that p+4-4x^2 >0, then why did they say that the solution they found was "**only**a solution" when this is true?posted by mhum at 3:13 PM on February 24, 2011

They say "if p+4 > 4x^2 then we can square both sides" because if it isn't, then the equation you have there doesn't make any sense. That is, "if this inequality doesn't hold, we stop, because then there is no value of x which would make this equation true because we would have the square root of a quantity being a negative number".

If you can change the sign on both roots simultaneously, why can't you change it on one and not the other, and have yet another case to consider?

The exposition is not the best and I think the author could have done a better job of leading the reader in the right direction, but that is the only way I think it can be interpreted.

posted by King Bee at 4:09 PM on February 24, 2011

If you can change the sign on both roots simultaneously, why can't you change it on one and not the other, and have yet another case to consider?

The exposition is not the best and I think the author could have done a better job of leading the reader in the right direction, but that is the only way I think it can be interpreted.

posted by King Bee at 4:09 PM on February 24, 2011

*If you can change the sign on both roots simultaneously, why can't you change it on one and not the other, and have yet another case to consider?*

I'm not sure I was clear on this point. I was implying that since for every solution x<0>0, we could wlog assume x>0. We couldn't change the sign of only one root because it wouldn't guarantee that we would preserve a solution.

Of course, all of this is moot. Upon closer inspection, I would have to agree that they don't seem to be considering the possibility of negative roots. The wording in the exposition is, in my opinion, really not great. I would have used "since" instead of "if" and "however".

It does appear that the only time people raise the possibility of a surd representing a negative root is only to rule it out (see also Question 2 in the 1990 USAMO). So, aside from a century-old text (classic though it may be), there really is no significant support for my previous position.

posted by mhum at 7:33 PM on February 24, 2011

This thread is closed to new comments.

posted by Green Eyed Monster at 9:01 PM on February 23, 2011