Question

Two candidates attempt to solve a quadratic equation of the form $x^2+px+q=0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$. The other starts with a wrong value of $q$ and finds the roots to be $-2$ and $9$. Find the sumof the correct roots of the equation.

Answer 1

In case $1$

Values of $p$ is wrong

So, $2+6\ne \frac{-b}{a}$

$8\ne \frac{-P}{1}$

$P\ne -8$

So, he find values of $2\times 6=\frac{q}{a}$ correctly

$12=q/1$

$q=12$

In case $2$

Here $q$ is worng

So, $-2\times q\ne q/a$

$⟹-18\ne q/1$

$⟹q\ne -18$

So, he finds value of $(-2)+9=-p/a$

$⟹7=-p/1$

$\therefore p=-7$

Thus, $x^2-7x+12=0$

$(x+3)(x-4)=0$

Sum of the roots = $-3+4$ = $1$