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Difference between revisions of "1989 AHSME Problems/Problem 8"

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This gives us 9 integers for <math>n</math>, <math>\boxed{\text{D}}</math>.
 
This gives us 9 integers for <math>n</math>, <math>\boxed{\text{D}}</math>.
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==Solution 2==
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For <math>x^2+x-n</math> to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals <math>\sqrt{1^2-4(1)(-n)}</math> or <math>\sqrt{4n+1}</math>.
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Since <math>n</math> must be a positive integer, <math>4n+1</math> must be odd. Therefore, <math>n</math> must be even.
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We find that there are 9 perfect squares that can be expressed in this way, while
  
 
== See also ==
 
== See also ==

Revision as of 13:53, 30 June 2018

Problem

For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$

Solution

For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$, where $a$ and $b$ are integers.

By expansion of the product of the linear factors and comparison to the original quadratic,

$ab = -n$

$a + b = 1$.

The only way for this to work if $n$ is a positive integer is if $a = -b +1$.

Here are the possible pairs:

$a = -1, b = 2$

$a = -2, b = 3$

$\vdots$

$a = -9, b = 10$

This gives us 9 integers for $n$, $\boxed{\text{D}}$.

Solution 2

For $x^2+x-n$ to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals $\sqrt{1^2-4(1)(-n)}$ or $\sqrt{4n+1}$.

Since $n$ must be a positive integer, $4n+1$ must be odd. Therefore, $n$ must be even.

We find that there are 9 perfect squares that can be expressed in this way, while

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AHSME Problems and Solutions

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