Difference between revisions of "2002 AMC 10A Problems/Problem 14"

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==Solution==
 
==Solution==
Consider a general quadratic with the coefficient of <math>x^2</math> as one and the roots as r and s. It can be factored as <math>(x-r)(x-s)</math> which is just <math>x^2-(r+s)x+rs</math>. Thus, the sum of the roots is the negative of the coefficient of x and the product is the constant term. (In general, this leads to Vieta's Formulas).
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Consider a general quadratic with the coefficient of <math>x^2</math> as one and the roots as r and s. It can be factored as <math>(x-r)(x-s)</math> which is just <math>x^2-(r+s)x+rs</math>. Thus, the sum of the roots is the negative of the coefficient of x and the product is the constant term. (In general, this leads to [[Vieta's Formulas]]).
  
 
We now have that the sum of the two roots is 63 while the product is k. Since both roots are primes, one must be 2, otherwise the sum is even. That means the other root is 61 and the product must be 122. Hence, our answer is <math>\boxed{\text{(B)}\ 1 }</math>.
 
We now have that the sum of the two roots is 63 while the product is k. Since both roots are primes, one must be 2, otherwise the sum is even. That means the other root is 61 and the product must be 122. Hence, our answer is <math>\boxed{\text{(B)}\ 1 }</math>.

Revision as of 18:50, 26 December 2008

Problem

The 2 roots of the quadratic $x^2 - 63x + k = 0$ are both prime. How many values of k are there?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E)}&$ (Error compiling LaTeX. Unknown error_msg)More than 4

Solution

Consider a general quadratic with the coefficient of $x^2$ as one and the roots as r and s. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of x and the product is the constant term. (In general, this leads to Vieta's Formulas).

We now have that the sum of the two roots is 63 while the product is k. Since both roots are primes, one must be 2, otherwise the sum is even. That means the other root is 61 and the product must be 122. Hence, our answer is $\boxed{\text{(B)}\ 1 }$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions