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### Quadratic equation real roots parameter range

Question:

The quadratic equation 2x2 - 5x + m = 0 has real roots. What is the range of m?

Solution:

The quadratic equation ax2 + bx + c = 0 has real roots when b2 - 4ac >= 0

Compare the standard quadratic equation ax2 + bx + c = 0 with the given quadratic equation 2x2 - 5x + m = 0. We get, a = 2, b= -5 and c = m

b2 - 4ac = (-5)2 - 4 × 2 m = 25 - 8m >= 0

8m <= 25
m <= 25/8

Case 1: when m = 25/8, the quadratic equation is: 2x2 - 5x + 25/8 = 0. We use the formula to find roots of the quadratic equation.

x1,2 = [-b +- square root of (b2 - 4ac)]/2a
= [-(-5) +- square root of ((-5)2 - 4 × 2 × 25/8)]/2 × 2
= [5 +- square root of (25 - 25)]/4
= 5/4

Therefore, when m = 25/8, x1 = x2 = 5/4

Case 2: now we choose a value of m which is less than 25/8. Choose m = -25/8.

x1,2 = [-b +- square root of (b2 - 4ac)]/2a
= [-(-5) +- square root of ((-5)2 - 4 × 2 × -25/8)]/2 × 2
= [-(-5) +- square root of ((-5)2 + 4 × 2 × 25/8)]/2 × 2
= [5 +- square root of (25 + 25)]/4
= [5 +- 5 square root of (1 + 1]/4
= (5 +- 5 square root of 2)/4
= (5/4) (1 +- square root of 2)

So, x1 = (5/4) (1 + square root of 2) and x2 = (5/4) (1 - square root of 2).

Therefore, when b2 - 4ac = 0, the quadratic equation has two equal real roots. When b2 - 4ac > 0, the quadratic equation has two different real roots. Watch the video for more details.