Prove that all the roots are real

[homeschool girl]

The problem:

"All the roots of
[math]x^2 + px + q = 0[/math]are real, where p and q are real numbers. Prove that all the roots of
[math]x^2 + px + q + (x + a)(2x + p) = 0[/math]are real, for any real number a."



What I've got so far:

[math]x^2 + px + q = 0[/math] gives the true statement of [math]p^2-4q \geq 0[/math].
We need to prove that [math]x^2 + px + q + (x + a)(2x + p) = 0[/math] has real roots and, se we need to prove the discriminant is non-negative.
writing it in standard form, I got [math]3x^2 + (2p + 2a)x +ap+q= 0[/math] and i got [math]4a^2+(-4p)a +(-4p^2-12q)[/math] as the discriminant, so the equation that needs to be proven is [math]4a^2+(-4p)a +(-4p^2-12q)\geq0[/math].
I'm not sure where to go from here except that I think I need to use [math]p^2-4q \geq 0[/math]

 

[lev888]

The problem:

"All the roots of
[math]x^2 + px + q = 0[/math]are real, where p and q are real numbers. Prove that all the roots of
[math]x^2 + px + q + (x + a)(2x + p) = 0[/math]are real, for any real number a."



What I've got so far:

[math]x^2 + px + q = 0[/math] gives the true statement of [math]p^2-4q \geq 0[/math].
We need to prove that [math]x^2 + px + q + (x + a)(2x + p) = 0[/math] has real roots and, se we need to prove the discriminant is non-negative.
writing it in standard form, I got [math]3x^2 + (2p + 2a)x +ap+q= 0[/math] and i got [math]4a^2+(-4p)a +(-4p^2-12q)[/math] as the discriminant, so the equation that needs to be proven is [math]4a^2+(-4p)a +(-4p^2-12q)\geq0[/math].
I'm not sure where to go from here except that I think I need to use [math]p^2-4q \geq 0[/math]

wolframalpha.com says the discriminant is 4(a^2 - ap + p^2 - 3q), so 4p^2 has wrong sign. What do we know about a?