How many real solutions?
The question
x² + 13x + 105 = 0
How many real solutions?
x² + 13x + 105 = 0
x² + 13x + 105 = 0
a = 1b = 13c = 105
b² − 4ac = 13² − 4(1)(105)
= −251 < 0→none
- You do not have to solve it. The discriminant, b² − 4ac — the part under the square root in the quadratic formula — tells you how many real solutions there are.
- Read off a, b and c from x² + 13x + 105 = 0: a = 1, b = 13, c = 105.
- Work it out: 13² is 169, and 4 × 1 × 105 is 420. That leaves 169 − 420 = −251.
- −251 is below zero, and no real number squares to a negative, so there is nothing to take the square root of. So there are NO real solutions.
How it works.
- 01
You do not have to solve it. The discriminant, b² − 4ac — the part under the square root in the quadratic formula — tells you how many real solutions there are.
- 02
Read off a, b and c from x² + 13x + 105 = 0: a = 1, b = 13, c = 105.
- 03
Work it out: 13² is 169, and 4 × 1 × 105 is 420. That leaves 169 − 420 = −251.
- 04
−251 is below zero, and no real number squares to a negative, so there is nothing to take the square root of. So there are NO real solutions.