x^2-7x+6>0 — step-by-step solution

Let's solve in

ℝ

x^{2} - 7 x + 6 > 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -7

and

c = 6

Δ = {(-7)}^{2} - 4 \cdot 1 \cdot 6

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

{(-7)}^{2}

(-7) \cdot (-7)

49

Δ = 49 - 4 \cdot 6

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 6

24

Δ = 49 - 24

ⓘ Start from

49

and move back

24

on the number line:

49 - 24

25

Δ = 25

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-7) - \sqrt{25}}{2 \cdot 1}

x_{1} = \frac{7 - \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{7 - 5}{2}

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

7 - 5

2

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

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

2

, their greatest common divisor:

\frac{2}{2}

1

x_{1} = 1

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-7) + \sqrt{25}}{2 \cdot 1}

x_{2} = \frac{7 + \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{7 + 5}{2}

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

7 + 5

12

x_{2} = \frac{12}{2}

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

2

, their greatest common divisor:

\frac{12}{2}

6

x_{2} = 6

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 (-∞, 1) \cup (6, +∞)