Answer to Question #302289 in Algebra for Carmen

Question

If the equation (1-q+(p^2)/2)x^2+p(1+q)x+q(q-1) +((p^2)/2)=0 has two equal roots, prove that p^2=4q

Answer 1

Quadratic equation has two equal roots if

$1-q+\dfrac{p^2}{2}\not=0$

$D=(p(1+q))^2-4(1-q+\dfrac{p^2}{2})(q(q-1)+\dfrac{p^2}{2})=0$

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

$p^2(1+2q+q^2)-(2-2q)(2q^2-2q)-p^2(2-2q)$

$-p^2(2q^2-2q+p^2)=0$

$-p^4+p^2(1+2q+q^2-2+2q-2q^2+2q)$

$-(2-2q)(2q^2-2q)=0$

$p^4+p^2(q^2-6q+1)-4q(q-1)^2=0$

If $p^2=4q$

$16q^2+4q^3-24q^2+4q-4q^3+8q^2-4q=0,$

True for $q\in \R.$

Or

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

The only solution is $p^2=0$ for $q=1.$

But in this case

$1-q+\dfrac{p^2}{2}=1-1+0=0$

and we have an equation

$0=0$

Hence the quadratic equation

$p^4+p^2(q^2-6q+1)-4q(q-1)^2=0$

has the only solution which satisfies our task

$p^2=4q.$

In this case the quadratic equation

$(1-q+\dfrac{p^2}{2})x^2+p(1+q)x+q(q-1)+\dfrac{p^2}{2}=0$

has two equal roots.