Question

Two candidates attempt to solve a quadratic equation $x^2+px+q=0$. One starts with a wrong value of $p$ and gets $2$, $6$ as its roots and other starts with wrong value of $q$ and obtains roots $2$, $-9$. The correct roots are

• A. $-3,-4$
• B. $3,4$
• C. $-3,4$
• D. $-4,3$

Answer 1

The correct option is A $-3,-4$

The quadratic equation with roots $2$ and $6$ is
$(x-2)(x-6)$
$x^2-8x+12$
Hence, $q=12$
Quadratic equation with roots $2,-9$ is
$(x-2)(x+9)$
$x^2+7x-18$
Hence, $p=7$
Therefore, the original equation is
$x^2+7x+12=0$
$(x+4)(x+3)=0$
$x=-4$ and $x=-3$.