Resolved: "Solve x6−7x3−8=0 algebraically. Find real solutions only, if..."

High School
Algebra
Algebra I
Algebra foundations

Kathleen Rausch

Answered question

2021-12-26

Solve

x^{6} - 7 x^{3} - 8 = 0

algebraically. Find real solutions only, if they exist.

Answer & Explanation

Elois Puryear

Beginner2021-12-27Added 30 answers

Put

u = x^{3}

then rewrite the equation

u^{2} - 7 u - 8 = 0

u^{2} - 8 u + u - 8 = 0

u (u - 8) + 1 (u - 8) = 0

(u + 1) (u - 8) = 0

u + 1 = 0, u - 8 = 0

u = - 1 or 8 u = 8

Put

u = x^{3}

then solve x

u = - 1 \Rightarrow x^{3} = - 1

\Rightarrow (x + 1) (x^{2} - x + 1) = 0

\Rightarrow x + 1 = 0 or x^{2} - x + 1 = 0

x = - 1 or x = \frac{+ 1 \pm \sqrt{1 - 4}}{2}

x = \frac{1 \pm \sqrt{- 3}}{2}

x = \frac{1 \pm \sqrt{3} i}{2}

[a x^{2} + b x + c = 0 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}]

and

u = 8

x^{3} = 8 \Rightarrow x^{3} = {(2)}^{3} \Rightarrow x^{2} - {(2)}^{3} = 0

\Rightarrow (x - 2) (x^{2} + 2 x + 4) = 0

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

\Rightarrow x - 2 = 0 or x^{2} + 2 x + 2 = 0

x = 2 or x = \frac{- 2 \pm \sqrt{4 - 16}}{2}

x = \frac{- 2 \pm \sqrt{- 12}}{2}

x = \frac{- 2 \pm \sqrt{12} i}{2}

x = \frac{- 2 \pm 2 \sqrt{3} i}{2}

x = - 1 \pm \sqrt{3} i

Hence real roots are -1 and 2.

Orlando Paz

Beginner2021-12-28Added 42 answers

$x^{6} - 7 x {3} - 8 = 0$
$x^{3} = t$
$t^{2} - 7 t - 8 = 0$
$D = 49 + 32 = 81$
$t 1 = \frac{7 + 9}{2} = 8$
$t 2 = \frac{7 - 9}{2} = - 1$
$x^{3} = 8$
$x^{3} = - 1$
$x 1 = 2$
$x 2 = - 1$

karton

Expert2022-01-04Added 613 answers

$\begin{matrix} x^{6} - 7 x^{3} - 8 = 0 \\ L e t P^{3} = a, t h e n \\ a^{2} - 7 a - 8 = 0 \\ 7 + - \sqrt{49} + 32 \\ a = ? \\ a = 8 \\ a = - 1 \\ x^{3} = 8 a n d x^{3} = - 1 \\ х = 2 х = - 1 \end{matrix}$
Answer: 2, -1

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