# Difference between revisions of "1983 AHSME Problems/Problem 18"

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Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as | Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as | ||

− | \begin{align*} | + | <math>\begin{align*} |

f(y) &= x^4 + 5x^2 + 3 \\ | f(y) &= x^4 + 5x^2 + 3 \\ | ||

&= (x^2)^2 + 5x^2 + 3 \\ | &= (x^2)^2 + 5x^2 + 3 \\ | ||

Line 15: | Line 15: | ||

&= y^2 - 2y + 1 + 5y - 5 + 3 \\ | &= y^2 - 2y + 1 + 5y - 5 + 3 \\ | ||

&= y^2 + 3y - 1. | &= y^2 + 3y - 1. | ||

− | \end{align*} | + | \end{align*}</math> |

Then substituting <math>x^2 - 1</math>, we get | Then substituting <math>x^2 - 1</math>, we get | ||

− | \begin{align*} | + | <math>\begin{align*} |

f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ | f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ | ||

&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ | &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ | ||

&= \boxed{x^4 + x^2 - 3}. | &= \boxed{x^4 + x^2 - 3}. | ||

− | \end{align*} | + | \end{align*}</math> |

The answer is (B). | The answer is (B). |

## Revision as of 14:55, 1 July 2017

Problem: Let be a polynomial function such that, for all real , For all real , is

(A) (B) (C) (D) (E) none of these

Solution:

Let . Then , so we can write the given equation as $\begin{align*} f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1. \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.) Then substituting , we get $\begin{align*} f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= \boxed{x^4 + x^2 - 3}. \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.) The answer is (B).