Question

If one root of $x^2+px+12=0$ is $4$ while the equation $x^2+px+q=0$ has equal roots, then the value of $q$ is

• A. $-7$
• B. $4$
• C. $42$
• D. $\frac{49}{4}$

Answer 1

The correct option is D $\frac{49}{4}$

If $4$ is a root of the equation $x^2+px+12=0$

This implies it satisfies the equation $x^2+px+12=0$, so we substitute $x=4$ as follows:

$x^2+px+12=0\Rightarrow 4^2+4p+12=0\Rightarrow 16+4p+12=0\Rightarrow 4p+28=0\Rightarrow 4p=-28\Rightarrow p=-\frac{28}{4}\Rightarrow p=-7$

Now, we substitute $p=-7$ in the other given equation $x^2+px+q=0$ as shown below:

$x^2+px+q=0\Rightarrow x^2-7x+q=0$

It is given that the equation $x^2+px+q=0$ or $x^2-7x+q=0$ has equal roots. This implies the discriminant $D=b^2-4ac$ must be equal to $0$. Therefore, we have:

$D=b^2-4ac\Rightarrow 0=(-7{)}^2-(4\times 1\times q)\Rightarrow 0=49-4q\Rightarrow 4q=49\Rightarrow q=\frac{49}{4}$

Hence, $q=\frac{49}{4}$.