How do you find all the real and complex roots of $x^{5} - 32 = 0$ $x^5 - 32 = 0$ ?

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How do you find all the real and complex roots of $x^{5} - 32 = 0$ $x^5 - 32 = 0$ ?

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Answer 1 · 2016-01-23T23:08:00

You must mean the solutions to the equation. In the solution below I will show you how to solve this equation

Explanation:

$x^{5}$ $x^5$ - 32 = 0

$x^{5}$ $x^5$ = 32

Now, since $x^{2}$ $x^2$ is opposite to √x, and $x^{3}$ $x^3$ is opposite to $\sqrt[3]{x}$ $root(3)x$ , $x^{5}$ $x^5$ must be opposite to $\sqrt[5]{x}$ $root(5)x$ .

x = $\sqrt[5]{32}$ $root(5)(32)$

x = 2

Since an odd exponent with a negative base always gives a negative number (example $- 2^{5}$ $-2^5$ = -32), x = 2 is the only possible solution to this equation.

Exercises:

Solve for x in the following equations.

a) $x^{3}$ $x^3$ - 27 = 0

b) $x^{4}$ $x^4$ - 16 = 0

Hopefully you understand now, and best of luck in your continuation.

Answer 2 · 2016-02-06T03:26:47

$x = 2, \frac{- 1 - \sqrt{5} \pm i \sqrt{10 - 2 \sqrt{5}}}{2}, \frac{\sqrt{5} - 1 \pm i \sqrt{10 + 2 \sqrt{5}}}{2}$ $x=2,(-1-sqrt5+-isqrt(10-2sqrt5))/2,(sqrt5-1+-isqrt(10+2sqrt5))/2$

Explanation:

It is evident that $x = 2$ $x=2$ is a root since $32 = 2^{5}$ $32=2^5$ . The real trick here is finding the remaining $4$ $4$ complex roots.

Since $x = 2$ $x=2$ is a root, $(x - 2)$ $(x-2)$ is a factor of the quintic. The remaining roots can be found through solving the remaining quartic:

$\frac{x^{5} - 32}{x - 2} = x^{4} + 2 x^{3} + 4 x^{2} + 8 x + 16$ $(x^5-32)/(x-2)=x^4+2x^3+4x^2+8x+16$

We want to solve for $x$ $x$ :

$x^{4} + 2 x^{3} + 4 x^{2} + 8 x + 16 = 0$ $x^4+2x^3+4x^2+8x+16=0$

I will admit that finding the complex roots is difficult and unintuitive (I resorted to using Google to find a way to factor this):

Notice that ${(x^{2} + x + 4)}^{2} = x^{4} + 2 x^{3} + 9 x^{2} + 8 x + 16$ $(x^2+x+4)^2=x^4+2x^3+9x^2+8x+16$ , which is almost the quartic we had remaining.

Thus, we can say that

$x^{4} + 2 x^{3} + 9 x^{2} + 8 x + 16 - 5 x^{2} = 0$ $x^4+2x^3+9x^2+8x+16-5x^2=0$

Which simplifies to be

${(x^{2} + x + 4)}^{2} - 5 x^{2} = 0$ $(x^2+x+4)^2-5x^2=0$

This can be factored as a difference of squares.

$(x^{2} + x + 4 + x \sqrt{5}) (x^{2} + x + 4 - x \sqrt{5}) = 0$ $(x^2+x+4+xsqrt5)(x^2+x+4-xsqrt5)=0$

These quadratic equations can be solved individually:

The first:

$x^{2} + (1 + \sqrt{5}) x + 4 = 0$ $x^2+(1+sqrt5)x+4=0$

$x = \frac{- 1 - \sqrt{5} \pm \sqrt{{(1 + \sqrt{5})}^{2} - 16}}{2}$ $x=(-1-sqrt5+-sqrt((1+sqrt5)^2-16))/2$

$x = \frac{- 1 - \sqrt{5} \pm \sqrt{- 10 + 2 \sqrt{5}}}{2}$ $x=(-1-sqrt5+-sqrt(-10+2sqrt5))/2$

$x = \frac{- 1 - \sqrt{5} \pm i \sqrt{10 - 2 \sqrt{5}}}{2}$ $x=(-1-sqrt5+-isqrt(10-2sqrt5))/2$

The second:

$x^{2} + (1 - \sqrt{5}) x + 4 = 0$ $x^2+(1-sqrt5)x+4=0$

$x = \frac{\sqrt{5} - 1 \pm \sqrt{{(1 - \sqrt{5})}^{2} - 16}}{2}$ $x=(sqrt5-1+-sqrt((1-sqrt5)^2-16))/2$

$x = \frac{\sqrt{5} - 1 \pm \sqrt{- 10 - 2 \sqrt{5}}}{2}$ $x=(sqrt5-1+-sqrt(-10-2sqrt5))/2$

$x = \frac{\sqrt{5} - 1 \pm i \sqrt{10 + 2 \sqrt{5}}}{2}$ $x=(sqrt5-1+-isqrt(10+2sqrt5))/2$

Answer 3 · 2016-02-07T23:33:46

$x = 2 e^{\frac{2 k π i}{5}}, k = 0, 1, 2, 3, 4$ $x=2e^((2kpii)/5), k=0,1,2,3,4$

Explanation:

$x^{5} - 32 = 0$ $x^5-32=0$

$x^{5} = 32$ $x^5=32$

$= \sqrt[5]{32}$ $=root(5)(32)$

The real solution is obviously 2.

Let's extend the equation to polar complexes:

$x^{5} = 32 = 32 e^{2 k π i}$ $x^5=32=32e^(2kpii)$ , with $k$ $k$ any integer, but pratically, from 0 to 4.

$x = \sqrt[5]{32 e^{2 k π i}}$ $x=root(5)(32e^(2kpii))$

$x = 2 e^{\frac{2 k π i}{5}}, k = 0, 1, 2, 3, 4$ $x=2e^((2kpii)/5), k=0,1,2,3,4$

MORE EXPLANATION OR ANOTHER WAY OF LOOKING AT IT:

An easy way to see this to imagine that the solution divided up around the circle of radius 2 in complex plane. Now divide the circle by 5 so the solutions will be: $x_{1} = 2 ∠ θ$ $x_1 = 2/_ theta$

1) Real $x_{1} = 2 ∠ (θ = 0)$ $x_1 = 2/_ (theta = 0)$
2) Complex $x_{2} = 2 ∠ (θ = \frac{2 π}{5})$ $x_2 = 2/_ (theta = (2pi)/5)$
Find the real and imaginary part: $2 cos θ + 2 i sin θ$ $2costheta + 2isintheta$
$R = 2 cos (\frac{2 π}{5}); I = sin (\frac{2 π}{5})$ $R = 2cos((2pi)/5); I = sin((2pi)/5)$
3) Complex $x_{3} = 2 ∠ (θ = \frac{4 π}{5})$ $x_3 = 2/_ (theta = (4pi)/5)$ this is $2 (\frac{2 π}{5})$ $2((2pi)/5)$
Find the real and imaginary part: $2 cos θ + 2 i sin θ$ $2costheta + 2isintheta$
4) Complex $x_{4} = 2 ∠ (θ = \frac{6 π}{5})$ $x_4 = 2/_ (theta = (6pi)/5)$ this is $3 (\frac{2 π}{5})$ $3((2pi)/5)$
Find the real and imaginary part: $2 cos θ + 2 i sin θ$ $2costheta + 2isintheta$
4) Complex $x_{5} = 2 ∠ (θ = \frac{8 π}{5})$ $x_5 = 2/_ (theta = (8pi)/5)$ this is $4 (\frac{2 π}{5})$ $4((2pi)/5)$
Find the real and imaginary part: $2 cos θ + 2 i sin θ$ $2costheta + 2isintheta$

Hope this help...

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How do you find all the real and complex roots of $x^{5} - 32 = 0$ $x^5 - 32 = 0$ ?

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How do you find all the real and complex roots of x^5 - 32 = 0x5−32=0x^5 - 32 = 0?

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How do you find all the real and complex roots of $x^{5} - 32 = 0$ $x^5 - 32 = 0$ ?