SOLUTION: For this problem: All the roots of x^2 + px + q = 0 are real, where p and q are real numbers. Prove that all the roots of x^2 + px + q + (x + a)(2x + p) = 0 are real, for any rea

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Question 1172141: For this problem: All the roots of
x^2 + px + q = 0 are real, where p and q are real numbers. Prove that all the roots of
x^2 + px + q + (x + a)(2x + p) = 0 are real, for any real number a.
I solved it like this:
We know that all roots of x^2+px+q=0 are real. We can derive from this condition that p^2- 4q >=0
Let us do some simplification of second equation:
x^2+ px + q (x+a)(2x+p) = 3x^2+ x(2p + 2a) + ap + q
So we want to prove that the discriminant of equation
3x^2+x(2p + 2a) + ap + q = 0
is greater or equal to zero.
D = (2p + 2a)^2 - 4 *3(ap+q) = 4(p^2 + a^2 + 2ap - 3ap - 3q) = 4(a^2-ap+p^2-3q)
To prove that D >=0 , we can view D as a polynomial of a:
D(a) = 4a^2 - a(4p) + 4(p^2-3q)
We know that if a quadratic polynomial has a positive greatest coefficient and it's discriminant <= 0 then polynomials are always positive.
So it remains for us to prove that
(4p)^2 - 4*4*4(p^2-3q) <= 0
Let's divide both side by 4^2 :
p^2 - 4(p^2-3q) <= 0
-3p^2 + 12q <= 0
Now we divide by 3
-p^2+4q<= 0
Now we transfer terms to the other side:
0<= p^2 -4q, or p^2-4q >=0
There, now we have proved that the roots of a are real!
Can you show me another way to solve it? I'm curious.
Found 4 solutions by Edwin McCravy, AnlytcPhil, Plocharczyk, mccravyedwin:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!

Here is a different way, borrowing from linear programming the idea of
using a non-negative slack variable to change an inequality into an equation.

Since  has only real solutions, 





Introduce a non-negative slack variable s  so that




Substitute for q in 

 



Which simplifies to this



Which has the discriminant



Which simplifies to



and it must be non-negative, so



We solve this for a

 

OR

 

But since a is real, this tells us that our
slack variable s, is not only non-negative but
it is also non-positive  as well.
This means s=0.

Therefore,  becomes 



becomes



which has solutions:

 and 

which are real.

Edwin
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!

Answer by Plocharczyk(17)   (Show Source): You can put this solution on YOUR website!

Answer by mccravyedwin(407)   (Show Source): You can put this solution on YOUR website!

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