Question

If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax²+bx+c=0 and -ax²+bx+c=0 has real roots.

Answer 1

it is given that, a , b, c ∈ R such that ac ≠ 0

we have to show that at least one of the equations ax² + bx + c = 0 and -ax² + bx + c = 0 has real roots.

let's find discriminant of both equations.

for ax² + bx + c = 0, D = b² - 4ac

for -ax² + bx + c = 0, D = b² + 4ac

now here there are four cases possible.

• case 1 : a > 0, b > 0 and b² < 4ac

then, only -ax² + bx + c has real roots.

• case 2 : a > , b > 0 and b² > 4ac

then both equations have real roots.

• case 3 : ac < 0 and b² < 4ac

then, only ax² + bx + c has real roots

• case 4 : ac < 0 and b² > 4ac

then both equations have real roots.

here we can see that in all four cases at least one of the equations has real roots.

[hence proved ]