Question

Find the value of '$p$' so that equation ${4x}^2-8px+9=0$ has roots whose difference is $4$.

• A. $\pm 3$
• B. $\pm \frac{2}{5}$
• C. $\pm \frac{5}{2}$
• D. None of these

Answer 1

Answer: (C)

$\pm \frac{5}{2}$

Given equation is: $4x^2-8ps+9=0$

$\therefore \alpha +\beta =\frac{8p}{4}=2p;\alpha \beta =\frac{9}{4}$

$\alpha +\beta =2p$

$\alpha -\beta =4$.........Given

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

$\Rightarrow {\left(4\right)}^2={\left(2p\right)}^2-4\times \frac{9}{4}$

$\Rightarrow 16=4p^2-9\Rightarrow 25=4p^2$

$\Rightarrow p^2=\frac{25}{4}$

$\Rightarrow p=\pm \frac{5}{2}$