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.