Question

lf the difference of the squares of the roots of equation $x^2-6x+q=0$ is $24$, then the value of $q$ is:

• A. $-7$
• B. $8$
• C. $5$
• D. $4$

Answer 1

Answer: (C)

$5$

Let $\alpha ,\beta$ are roots of $x^2-6x+q=0,$ then

$S_1=\alpha +\beta =6$

And $S_2=\alpha \beta =q$

Given ${\alpha }^2-{\beta }^2=24$

Now from ${(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta$

$\Rightarrow {(\alpha -\beta )}^2=36-4q\Rightarrow (\alpha -\beta )=\sqrt{36-4q}$

As ${\alpha }^2-{\beta }^2=24\Rightarrow (\alpha -\beta )(\alpha +\beta )=24$

$\Rightarrow \sqrt{36-4q}\left(6\right)=24\Rightarrow \sqrt{36-4q}=4$

$\Rightarrow 36-4q=16\Rightarrow 4q=20\Rightarrow q=5$