Question

The value of $p$ for which one root of the quadratic equation $(p^2-5p+3)x^2+(3p-1)x+2=0$ is twice as large as the other is

• A. $-\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $-\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Let $y$ and $2y$ be the roots of the quadratic equation $(p^2-5p+3)x^2+(3p-1)x+2=0$
Sum of roots: $y+2y=\frac{-3p+1}{p^2-5p+3}\Rightarrow y=\frac{-3p+1}{3(p^2-5p+3)}$ ....(1)
Product of roots: $2y^2=\frac{2}{p^2-5p+3}$ ....(2)
Substituting (1) in (2), we get
${\left(\frac{-3p+1}{3(p^2-5p+3)}\right)}^2=\frac{1}{p^2-5p+3}$
$\Rightarrow 9p^2-6p+1=9p^2-45p+27$
$\Rightarrow p=\frac{2}{3}$
Ans: D