Step by step solution to "Solve, please: (u−2)3+40=0, where u is a real..."

High School
Algebra
Algebra I
Exponents and radicals

Jackson Floyd

Answered question

2022-03-21

Solve, please:

{(u - 2)}^{3} + 40 = 0

, where u is a real number

Answer & Explanation

clarkchica44klt

Beginner2022-03-22Added 17 answers

${(u - 2)}^{3} + 40 = 0$
Use odd root properl. If n is odd positive integer, then for any real number k $x^{n} = k$ is equivalent to $x = \sqrt[n]{k}$
Thus, put foirst $u - 2 = x$
And we get $x = (40)^{1 / 3} = \sqrt[3]{40}$
Convert given number to polar form
$40 = 40 i \sin (0) + 40 \cos (0)$
All n-th roots of complex number:
$γ ({\cos α}_{i} \sin α)$ are given by
$\sqrt[n]{γ} (\cos (\frac{α + 2 π k}{n}) + i \sin (\frac{α + 2 π k}{n}))$
$k = 0, 1, 2, 3, \dots, n - 1, n,$
We know tthat $r = 40, α = 0, n = 3$
Thus,
1) $k = 0$
$\sqrt[3]{40} (\cos (\frac{0 + 2 π (0)}{3}) + i \sin (\frac{0 + 2 π (0)}{3}))$
$= 2 \sqrt[3]{5} (\cos 0 + i \sin 0) = 2 \sqrt[3]{5}$
2) $k = 1$
$\sqrt[3]{40} (\cos (\frac{0 + 2 π (1)}{3}) + i \sin (\frac{0 + 2 π (1)}{3}))$
$= 2 \sqrt[3]{5} (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$
$= - \sqrt[3]{5} + \sqrt{2} \sqrt[3]{5 i}$
3) $k = 2$
$\sqrt[3]{40} (\cos (\frac{0 + 2 π (2)}{3}) + i \sin (\frac{0 + 2 π (2)}{3}))$
$= 2 \sqrt[3]{5} (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$
$= - \sqrt[3]{5} - \sqrt{3} \sqrt[3]{5 i}$
Thus, roots: $x_{1} = 2 \sqrt[3]{5}$
$x_{2} = - \sqrt[3]{5} + \sqrt{3} \sqrt[3]{5 i}$
$x_{3} = - \sqrt[3]{5} - \sqrt{3} \sqrt[3]{5 i}$
However, we have $u - 2 = x$
So add $- 2$ on both sides, and we get
$u_{1} = 2 (\sqrt[3]{5} - 1)$
$u_{2} = - \sqrt[3]{5} - 2 + \sqrt{3} \sqrt[3]{5 i}$
$u_{2} = - \sqrt[3]{5} - 2 - \sqrt{3} \sqrt[3]{5 i}$

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