Question

If a, b, c are real numbers such that ac is not equal to zero, show that at least one of the equations
1) ax^2 + bx + c = 0, and
2) -a(x)^2 + bx + c = 0,
has real roots.

Answer 1

You can also show it has non real roots just use values that make them real.

Answer 2

But, we have to prove it, use discrimant to decide signs, and show the result.

Answer 3

discriminant1 = (b^2 -4ac)
discriminant2 = (b^2 +4ac)


to have real roots one of those must be greater or equal to 0
b^2 -4ac > 0 b^2 +4ac > 0
b^2 > 4ac b^2 > -4ac

however this does not prove that either of them are true for any a,b,c

Answer 4

wait let me continue

b^2 has to be positive, ac can be either positive or negative

b^2/ac >4 or b^2/ac > -4
if ac is positive, then b^2/ac > -4 must be true
if ac is negative, then b^2/ac >4 must be false

Answer 5

This question's already been answered. If ac>0, b^2/ac>0 then b^2>4ac is possible and the first equation may have real roots. In this case, the second equation will have real roots also, since it satisfies b^2/ac>-4
If ac<0, b^2/ac<0 and the first equation wont have real roots. But the second equation will have real roots if -4<(b^2/ac)<0(if b^2/ac lies in between 0 and -4)