Question

If $p\in Q$ and the quadratic equations $x^2-4x+1=0$ and $p{x}^2-(p^2+3)x+2p^2-p=0$ have a root in common then the value (s) of p are

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $1$

Given: $x^2-4x+1=0$

Roots of $x^2-4x+1$ is $x=\frac{4\pm \sqrt{16-4}}{2}$

$x=2+\sqrt{3},2-\sqrt{3}$

Also, given that $p{x}^2-(p^2+3)x+2p^2-p$ have

same root as $x^2-4x+1$. So,

product of roots $=\frac{c}{2}$

here, $c=2p^2-p$ and $a=p$.

$(2+\sqrt{3})(2-\sqrt{3})=\frac{2p^2-p}{p}$ p is not equal to 0.

$4-3=\frac{2p^2-p}{p}$

$2p^2-p=p$

$2p^2-2p=0$

$2p(p-1)=0$

$p=0$ and $p=1$

But we consider only p=1

Option (A) is correct.