Question

Let $\alpha$ and $\beta$ are the roots of equation$2x^2-(p+1)x+(p-1)=0$. If $\alpha -\beta =\alpha \beta$, then what is the value of $p$ ?

• A. $1$
• B. $2$
• C. $3$
• D. $-2$

Answer 1

The correct option is B

$2$

Finding the value of $p$

Sum of roots $\alpha$ and $\beta$

$\Rightarrow \alpha +\beta =\frac{\left(p+1\right)}{2}\left[\because \text{sum of roots of}a{x}^2+bx+c=0is\frac{-b}{a}\right]$

Product of roots $\alpha$ and $\beta$

$\Rightarrow \alpha \beta =\frac{\left(p-1\right)}{2}\left[\because \text{product of roots of}a{x}^2+bx+c=0is\frac{c}{a}\right]$

It is given that,

$$\begin{gathered} {(\alpha –\beta )}^2={\alpha }^2{\beta }^2 \\ \Rightarrow {{\alpha }^2+\beta }^2-2\alpha \beta ={\alpha }^2{\beta }^2\left[\because {\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ \Rightarrow {{\alpha }^2+\beta }^2-2\alpha \beta +4\alpha \beta ={\alpha }^2{\beta }^2+4\alpha \beta [\mathrm{Adding}4\mathrm{\alpha \beta }\mathrm{both}\mathrm{sides}] \\ \Rightarrow {{\alpha }^2+\beta }^2+2\alpha \beta ={\alpha }^2{\beta }^2+4\alpha \beta \\ \Rightarrow {(\alpha +\beta )}^2–4\alpha \beta ={\alpha }^2{\beta }^2\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \Rightarrow {\left(\frac{p+1}{2}\right)}^2–4\left(\frac{p-1}{2}\right)={\left(\frac{p-1}{2}\right)}^2 \\ \Rightarrow {\left(\frac{p+1}{2}\right)}^2–{\left(\frac{p-1}{2}\right)}^2=4\left(\frac{p-1}{2}\right) \\ \Rightarrow \left[\left(\frac{p+1}{2}\right)+\left(\frac{p-1}{2}\right)\right]\left[\left(\frac{p+1}{2}\right)-\left(\frac{p-1}{2}\right)\right]=4\left(\frac{p-1}{2}\right)\left[\because a^2-b^2=(a-b)(a+b)\right] \\ \Rightarrow p\times 1=2(p-1) \\ \Rightarrow p=2 \end{gathered}$$

Hence, option B is correct .