Question

If one root of $x^2+px-10=0$ is $2$ while the equation $x^2-px+q=0$ has equal roots, then the value of $q$ is

• A. $\frac{9}{16}$
• B. $9$
• C. $\frac{4}{9}$
• D. $\frac{9}{4}$

Answer 1

The correct option is D $\frac{9}{4}$

If $2$ is a root of the equation $x^2+px-10=0$

This implies it satisfies the equation $x^2+px-10=0,$ so we substitute $x=2$ as follows:

$x^2+px-10=0$

$\Rightarrow 2^2+2p-10=0$

$\Rightarrow 4+2p-10=0$

$\Rightarrow 2p-6=0$

$\Rightarrow 2p=6$

$\Rightarrow p=3$

Now we substitute $p=3$ in the other given equation $x^2-px+q=0$ as shown below:

$x^2-px+q=0$

$\Rightarrow x^2-3x+q=0$

It is given that the equation $x^2-px+q=0$ i.e, $x^2-3x+q=0$ has equal roots. This implies the discriminant $D=b^2-4ac$ must be equal to $0$. Therefore, we have:

$D=b^2-4ac$

$\Rightarrow 0=(-3{)}^2-(4\times 1\times q)$

$\Rightarrow 0=9-4q$

$\Rightarrow 4q=9$

$\Rightarrow q=\frac{9}{4}$

Hence, $q=\frac{9}{4}$.