Question

Find the value of $p$ for which the quadratic equation $x^2+p(4x+p-1)+2=0$ has equal roots ?

• A. $-1,\frac{2}{3}$
• B. $3,5$
• C. $-1,\frac{4}{3}$
• D. $\frac{3}{4},2$

Answer 1

The correct option is A $-1,\frac{2}{3}$

Given equation is $x^2+p(4x+p-1)+2=0$.
Comparing it with the $a{x}^2+bx+c$, we get

$a=1,b=4p,c=(p^2-p+2)$
Therefore, $D=b^2-4ac$
$x^2+p(4x+p-1)+2=0$.
$\Rightarrow x^2+4xp+p^2-p+2=0$
$\Rightarrow (4p{)}^2-4(p^2-p+2)=0$
$\Rightarrow 16p^2-4p^2+4p-8=0$
$\Rightarrow 12p^2+4p-8=0$
$\Rightarrow 3p^2+p-2=0$
$\Rightarrow 3p^2+3-2p-2=0$
$\Rightarrow 3p(p+1)-2(p+1)=0$
$\Rightarrow (p+1)(3p-2)=0$
$\Rightarrow p=-1$and $p=\frac{2}{3}$