6x^5 + 8x=26x^3 — step-by-step solution

1) Let's solve in

ℝ

6 x^{5} + 8 x = 26 x^{3}

You can write the expression:

6 x^{5} - 26 x^{3} + 8 x

in factored form:

x \cdot (6 x^{4} - 26 x^{2} + 8)

A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥1.1) Let's solve in

ℝ

x = 0

⇥Solution obtained:

x = 0

⇥1.2) Let's solve in

ℝ

6 x^{4} - 26 x^{2} + 8 = 0

⇥Since

2 \neq 0

, divide both sides by

2

3 x^{4} - 13 x^{2} + 4 = 0

⇥Let's define:

X = x^{2}

⇥You can transform this expression into:

3 X^{2} - 13 X + 4 = 0

⇥Given the quadratic form

a X^{2} + b X + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 3

b = -13

and

c = 4

⇥

Δ = {(-13)}^{2} - 4 \cdot 3 \cdot 4

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

{(-13)}^{2}

(-13) \cdot (-13)

169

■Multiplications are done before additions and subtractions:

3 \cdot 4

12

⇥

Δ = 169 - 12 \cdot 4

⇥ⓘ Multiplications are done before additions and subtractions:

12 \cdot 4

(10 + 2) \cdot 4

10 \cdot 4 + 2 \cdot 4

40 + 8

48

⇥

Δ = 169 - 48

⇥ⓘ Start from

169

and move back

48

on the number line:

169 - 48

121

⇥

Δ = 121

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

X_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-13) - \sqrt{121}}{2 \cdot 3}

⇥

X_{1} = \frac{13 - \sqrt{121}}{2 \cdot 3}

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

\sqrt{121}

11

;

11^{2}

121

⇥ⅰ

2 \cdot 3

6

⇥

X_{1} = \frac{13 - 11}{6}

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

13 - 11

2

⇥

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

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

2

, their greatest common divisor:

\frac{2}{6}

\frac{1 \cdot 2}{3 \cdot 2}

\frac{1}{3}

⇥

X_{1} = \frac{1}{3}

⇥The quadratic formula gives

X_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-13) + \sqrt{121}}{2 \cdot 3}

⇥

X_{2} = \frac{13 + \sqrt{121}}{2 \cdot 3}

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

\sqrt{121}

11

;

11^{2}

121

⇥ⅰ

2 \cdot 3

6

⇥

X_{2} = \frac{13 + 11}{6}

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

13 + 11

24

⇥

X_{2} = \frac{24}{6}

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

6

, their greatest common divisor:

\frac{24}{6}

4

⇥

X_{2} = 4

⇥Solutions obtained:

X = {\frac{1}{3}, 4}

⇥Let's solve in

ℝ

x^{2} = X = \frac{1}{3}

⇥The two numbers whose square is

\frac{1}{3}

are

- \frac{\sqrt{3}}{3}

and

\frac{\sqrt{3}}{3}

⇥Solutions obtained:

x = {- \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}}

⇥Let's solve in

ℝ

x^{2} = X = 4

⇥The two numbers whose square is

4

are

-2

and

2

⇥Solutions obtained:

x = {-2, 2}

▶Solution:

x = {-2, - \frac{\sqrt{3}}{3}, 0, \frac{\sqrt{3}}{3}, 2}

2) The numeric value of:

x = {-2, - \frac{\sqrt{3}}{3}, 0, \frac{\sqrt{3}}{3}, 2}

is:

x = {-2, \approx - \frac{1.7321}{3}, 0, \approx \frac{1.7321}{3}, 2}

▷

x = {-2, \approx -0.5774, 0, \approx 0.5774, 2}