# Difference between revisions of "2005 AMC 10A Problems/Problem 10"

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Alternatively, note that whatever the two values of <math>a</math> are, they must lead to equations of the form <math>px^2 + qx + r =0</math> and <math>px^2 - qx + r = 0</math>. So the two choices of <math>a</math> must make <math>a_1 + 8 = q</math> and <math>a_2 + 8 = -q</math> so <math>a_1 + a_2 + 16 = 0</math> and <math>a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}</math>. | Alternatively, note that whatever the two values of <math>a</math> are, they must lead to equations of the form <math>px^2 + qx + r =0</math> and <math>px^2 - qx + r = 0</math>. So the two choices of <math>a</math> must make <math>a_1 + 8 = q</math> and <math>a_2 + 8 = -q</math> so <math>a_1 + a_2 + 16 = 0</math> and <math>a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}</math>. | ||

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+ | ==Alternate Solution== | ||

+ | |||

+ | Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have | ||

+ | <cmath> (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 144. </cmath> We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the <math>\pm</math> sign when added). So we must have | ||

+ | <cmath> \frac{-16 \pm \sqrt{\text{something}}{2} + \frac{-16 \pm \sqrt{something}}{2}. </cmath> | ||

+ | Therefore, we have <math>(-16)(2)/2 = -16 \implies \boxed{A}.</math> | ||

==See Also== | ==See Also== |

## Revision as of 15:55, 21 December 2009

## Problem

There are two values of for which the equation has only one solution for . What is the sum of those values of ?

## Solution

A quadratic equation has exactly one root if and only if it is a perfect square. So set

Two polynomials are equal only if their coefficients are equal, so we must have

or .

So the desired sum is

Alternatively, note that whatever the two values of are, they must lead to equations of the form and . So the two choices of must make and so and .

## Alternate Solution

Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the sign when added). So we must have

\[\frac{-16 \pm \sqrt{\text{something}}{2} + \frac{-16 \pm \sqrt{something}}{2}.\] (Error compiling LaTeX. ! File ended while scanning use of \frac .)

Therefore, we have