Understanding Quadratic Equations: Real vs. Imaginary Solutions

If b²-4ac < 0

There are no roots

If b²-4ac < 0, then the quadratic equation ax²+bx+c=0 has no real solutions. This is because when b²-4ac is negative, the square root of a negative number is involved, which is not defined in the set of real numbers. For example, consider the quadratic equation x²+4x+5=0. Here, a=1, b=4, and c=5. Using the quadratic formula, we have: x = (-b ± √(b²-4ac)) / (2a) x = (-4 ± √(4²-4(1)(5))) / (2*1) x = (-4 ± √(-4)) / 2 Since the square root of a negative number is involved, we cannot find real solutions to this equation. Instead, we can express the solutions in terms of imaginary numbers. We write: x = (-4 ± 2i) / 2 x = -2 ± i Therefore, if b²-4ac < 0, the quadratic equation has no real solutions, but it may have complex or imaginary solutions.

More Answers: