The nature of quadratic equation

Question

Given that $p, q$ are real numbers. Prove that if the equation $2x^2+2(p+q)x+p^2+q^2=0$ have real roots, $p$ must be equal to $q$.

What I tried:

If the equation has equal roots, then it can be rewritten as: $$2x^2 + 4px + 2p^2 = 0$$ Coefficients of the quadratic equation are: $a=2$, $b=4p$, $c=2p^2$. If the quadratic equation has equal roots than the determinant value must be zero. $$16p^2 - 16p^2 = 0 $$

Here I am coming to a dead end. What am I missing?

Answer 1

The discriminant of $2x^2+2(p+q)x+p^2+q^2$, which is $4(p+q)^2-8(p^2+q^2)$,

must be non-negative for the quadratic to have real roots.

I.e., we need $(p+q)^2\ge2(p^2+q^2)$ or $2pq\ge p^2+q^2$ or $0\ge (p-q)^2$.

But $(p-q)^2\le0$ only when $(p-q)^2=0$, which happens only when ...

Answer 2

You didn't quite have the right start. All real roots are covered by the ABC formula. For $2x^2+2(p+q)x+p^2+q^2=0$ we have $a = 2$, $b = 2(p + q)$ and $c = p^2 + q^2$.

The discriminant of this equation is $D = b^2 - 4ac$, or $D = 4(p + q)^2 - 8(p^2 + q^2)$. A quadratic equation only has real roots when $D \geq 0$. We have:

$$4(p + q)^2 - 8(p^2 + q^2) \geq 0$$ $$(p + q)^2 \geq 2(p^2 + q^2)$$

Can you take it from here?

Answer 3

Having equal roots has very little to do with having real roots (it turns out it does here, but you don't know that yet!). Instead, complete the square: $$ 2x^2 + 2(p+q)x + p^2 + q^2 = 2\left( x + \frac{p+q}{2} \right)^2 + p^2+q^2 - \frac{(p+q)^2}{2} \\ = 2\left( x + \frac{p+q}{2} \right)^2 + \frac{p^2-2pq+q^2}{2} \\ = 2\left( x + \frac{p+q}{2} \right)^2 + \frac{(p-q)^2}{2} . $$ If $x$ is real, this is a sum of squares of real numbers, so can only be zero if both of the numbers inside the squares are zero, i.e. $p=q$ and $x=-(p+q)/2=-p$.