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
Question 1172144: 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
For this problem, is there any other ways to solve it? If so, can you show me? Thanks! Answer by Edwin McCravy(20055) (Show Source):
