Question

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

• A. $3$
• B. $12$
• C. $\frac{49}{4}$
• D. $4$

Answer 1

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

$x^2+px+12=0$ has one root as $4$.

Hence, let another root be $y$.

Then,

$4\left(y\right)=12$

$y=3$.

Hence, $p=4+3=7$

Now,

The other equation becomes

$x^2+7x+q=0$

Since it has equal roots,

$b^2=4ac$

$49=4q$

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