x^2-11x+24>0 — step-by-step solution

Let's solve in

ℝ

x^{2} - 11 x + 24 > 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -11

and

c = 24

Δ = {(-11)}^{2} - 4 \cdot 1 \cdot 24

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-11)}^{2}

(-11) \cdot (-11)

121

Δ = 121 - 4 \cdot 24

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 24

4 \cdot (20 + 4)

4 \cdot 20 + 4 \cdot 4

80 + 16

96

Δ = 121 - 96

ⓘ Start from

121

and move back

96

on the number line:

121 - 96

25

Δ = 25

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-11) - \sqrt{25}}{2 \cdot 1}

x_{1} = \frac{11 - \sqrt{25}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

x_{1} = \frac{11 - 5}{2}

ⓘ The numerator and the denominator are computed before the division:

11 - 5

6

x_{1} = \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

x_{1} = 3

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-11) + \sqrt{25}}{2 \cdot 1}

x_{2} = \frac{11 + \sqrt{25}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{25}

5

;

5^{2}

25

x_{2} = \frac{11 + 5}{2}

ⓘ The numerator and the denominator are computed before the division:

11 + 5

16

x_{2} = \frac{16}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{16}{2}

8

x_{2} = 8

The quadratic has the sign of a =

1

outside its roots, so it is indeed positive outside its roots; the roots, where it is zero, are excluded.

▶Solution found:

x \in (-∞, 3) \cup (8, +∞)