Question

Let p and q be real numbers such that the roots of
x^2 + px + q = 0 are real. Prove that the roots of
x^2 + px + q + (x + a)(2x + p) = 0 are real, for any real number a.

Answer 1

Answer:

Step-by-step explanation:

Known fact: all roots of quadratic equation $ax^2 + bx + c = 0$ are real if and only if

discriminant $D = b^2 - 4ac \geq 0.$

We know that all roots of $x^2 + px + q = 0$ are real. We can derive from this condition that

$p^2 - 4q \geq 0$. (<----------- HERE)

Let's do some simplification of second equation:

$x^2 + px +q + (x + a)(2x + p) = 3x^2 + x(2p + 2a) + ap + q.$

So we want according to Known fact prove that discriminant of equation

$3x^2 + x(2p + 2a) + ap + q = 0$ is greater or equal zero.

$D = (2p + 2a)^2 - 4\cdot 3 (ap + q) = 4(p^2 + a^2 + 2ap - 3ap - 3q) = 4(a^2 - ap + p^2 - 3q)$.

To prove that $D \geq 0$, we can view D as polynomial of a:

$D(a) = 4a^2 - a(4p) + 4(p^2 - 3q)$.

Known fact #2: if quadratic polynomial have positive greatest coefficient and it's discriminant $\leq 0$ then polynomial always positive.

So remains to prove that

$(4p)^2 - 4\cdot 4\cdot 4(p^2 - 3q) \leq 0$  

lets divide both side by $4^2$:

$p^2 - 4(p^2 - 3q) \leq 0$

$-3p^2 + 12q \leq 0$

divide by 3

$-p^2 + 4q \leq 0$

transfer terms to the other side:

$0 \leq p^2 - 4q$

But we know that from "(<----------- HERE)"!!!

So Q.E.D.