Question

If the roots of the equation
$(1-q+\frac{p^2}{2})x^2+p(1+q)x+q(q-1)+\frac{p^2}{2}=0$ are equal, then

• A. $p^2=8q$
• B. $p^2=2q$
• C. $p^2=4q$
• D. $p^2=q$

Answer 1

The correct option is C $p^2=4q$

As roots are equal, so $\Delta =0$
$\Rightarrow p^2(1+q{)}^2=4(1-q+\frac{p^2}{2})\left(q\right(q-1)+\frac{p^2}{2})$

$\Rightarrow p^2(1+q{)}^2=\left(4\right(1-q)+2p^2)\left(q\right(q-1)+\frac{p^2}{2})$

$\Rightarrow p^2(1+q{)}^2=-4q(1-q{)}^2+2p^{2}q(q-1)+2p^2(1-q)+p^4$

$\Rightarrow p^2(1+q{)}^2-2p^{2}q(q-1)-2p^2(1-q)-p^4=-4q(1-q{)}^2$

$\Rightarrow p^2\left[\right(1+q{)}^2-2q^2+4q-2-p^2]=-4q(1-q{)}^2$

$\Rightarrow p^2[-(1-q{)}^2+(4q-p^2)]=4q(1-q{)}^2$

$\Rightarrow -p^2(1-q{)}^2+p^2(4q-p^2)=-4q(1-q{)}^2$

$\Rightarrow (1-q^2)(-p^2+4q)+p^2(4q-p^2)=0$

$\Rightarrow (-p^2+4q)\left[\right(1-q{)}^2+p^2]=0$

As $(1-q{)}^2+p^2\ne 0$,
$-p^2+4q=0$

$\therefore p^2=4q$