Question 1172098: So for this question(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 did these steps:
all roots of quadratic equation ax^2+bx+c =0 are real if and only if
discriminant D = b^2- 4ac >= 0
I know that all roots of x^2+px+q=0 are real. We can derive from this condition that p^2- 4q >=0
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 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 polynomial of a:
.
D(a) = 4a^2 - a(4p) + 4(p^2-3q) if quadratic polynomial have positive greatest coefficient and it's discriminant <= 0 then polynomials are always positive.
So remains 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
divide by 3
-p^2+4q<= 0
transfer terms to the other side:
0<= p^2 -4q
Are these steps correct? Would there be another approach to solving this? If there is, what is it? Thanks.
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
The logic is correct.
My congratulations (!) --- you got really good solution.
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At which high school / college / university did you get this problem ?
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