Question

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and finds $-9$ and $-1$ for the roots. The correct equation was:

$x^2-10x+9=0$

$x^2+10x+9=0$

$x^2-10x+16=0$

$x^2-8x-9=0$

none of these

Answer: A

Concepts:quadratic roots Vieta formulas coefficients

Difficulty rating: 1660

Small Hint:

The first student’s correct linear coefficient is determined by the sum $8+2$

Big Hint:

The second student’s correct constant term is determined by the product $(-9)(-1)$

Solution:

The first student changed only the constant term, so the correct coefficient of $x$ is the one in the monic quadratic with roots $8$ and $2:$ it is $-(8+2)=-10.$ The second student changed only that linear coefficient, so the correct constant is $(-9)(-1)=9.$ Therefore the correct equation is $x^2-10x+9=0.$

Thus, the correct answer is A.

1953 AMC 12 Problem 44

Answer: A

Difficulty rating: 1660

Small Hint:

Big Hint:

Solution:

Problem 44 in Other Years