Prove that all the roots are real

homeschool girl

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Feb 6, 2020
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The problem:

"All the roots of
x2+px+q=0x^2 + px + q = 0are real, where p and q are real numbers. Prove that all the roots of
x2+px+q+(x+a)(2x+p)=0x^2 + px + q + (x + a)(2x + p) = 0are real, for any real number a."



What I've got so far:

x2+px+q=0x^2 + px + q = 0 gives the true statement of p24q0p^2-4q \geq 0.
We need to prove that x2+px+q+(x+a)(2x+p)=0x^2 + px + q + (x + a)(2x + p) = 0 has real roots and, se we need to prove the discriminant is non-negative.
writing it in standard form, I got 3x2+(2p+2a)x+ap+q=03x^2 + (2p + 2a)x +ap+q= 0 and i got 4a2+(4p)a+(4p212q)4a^2+(-4p)a +(-4p^2-12q) as the discriminant, so the equation that needs to be proven is 4a2+(4p)a+(4p212q)04a^2+(-4p)a +(-4p^2-12q)\geq0.
I'm not sure where to go from here except that I think I need to use p24q0p^2-4q \geq 0
 
The problem:

"All the roots of
x2+px+q=0x^2 + px + q = 0are real, where p and q are real numbers. Prove that all the roots of
x2+px+q+(x+a)(2x+p)=0x^2 + px + q + (x + a)(2x + p) = 0are real, for any real number a."



What I've got so far:

x2+px+q=0x^2 + px + q = 0 gives the true statement of p24q0p^2-4q \geq 0.
We need to prove that x2+px+q+(x+a)(2x+p)=0x^2 + px + q + (x + a)(2x + p) = 0 has real roots and, se we need to prove the discriminant is non-negative.
writing it in standard form, I got 3x2+(2p+2a)x+ap+q=03x^2 + (2p + 2a)x +ap+q= 0 and i got 4a2+(4p)a+(4p212q)4a^2+(-4p)a +(-4p^2-12q) as the discriminant, so the equation that needs to be proven is 4a2+(4p)a+(4p212q)04a^2+(-4p)a +(-4p^2-12q)\geq0.
I'm not sure where to go from here except that I think I need to use p24q0p^2-4q \geq 0
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?
 
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