We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Help me asap

0
719
5

There are six different sixth roots of 64. That is, there are six complex numbers that solve $x^6=64.$Find them.

Feb 3, 2018

### 5+0 Answers

#1
+1

Solve for x:

x^6 = 64

Take 6^th roots of both sides.

Taking 6^th roots gives 2 times the 6^th roots of unity:

x = -2    or x = 2     or x = -2 (-1)^(1/3)     or x = 2 (-1)^(1/3)    or x = -2 (-1)^(2/3)    or x = 2 (-1)^(2/3)

Feb 3, 2018
edited by Guest  Feb 3, 2018
#2
+1

x^6  =  64

64  =  64 + 0i

The sixth root divides the unit  circle into six even parts  and the beginning angle is

arctan (0 / 64)  =  0°

We have

6√64  [  cos (0) + i sin (0)]  =   2 [ 1 + 0 i ]  =  2 + 0i

6√64 [  cos (pi/ 3)  + i  sin (pi /3)  = 2 [ 1/2 + √3/2 i ]  =  1 + √3 i

6√64 [  cos (2pi/ 3) + i sin (2pi/ 3)  ]  = 2 [ -1/2  +  √3 i ]  =   - 1 + √3 i

6 √64 [  cos (pi)  + i sin (pi) ]   =     2 [ - 1 + 0 i ]  =  2[-1]  =  -2 + 0i

6√64  [ cos (4pi/3 + i sin (4pi/3) ]  =  2 [ -1/2  - √3 / 2 i ]  =  - 1 - √3 i

6√64 [ cos (5 pi / 3) + i sin (5pi/ 3) ]  = 2 [ 1/2  - i √3 /2 ]  =  1 - √3 i   Feb 3, 2018
edited by CPhill  Feb 3, 2018
#3
+1

It is also possible to use knowledge of difference and sum of squares and cubes to find all the possible solutions.

$$x^6=64$$   Move 64 to the other side.
$$x^6-64=0$$   This can be written as a difference of squares.
$$\left(x^3\right)^2-8^2=0$$   Since this is a difference of squares, factor it as such.
$$(x^3+8)(x^3-8)=0$$   Notice how both binomials are now difference and sum of cubes. Factor both completely.
$$(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0$$   Now, set each factor equal to zero and solve. Let's do the easy ones first.
 $$x+2=0$$ $$x-2=0$$ $$x_1=-2$$ $$x_2=2$$

Now, let's tackle the harder ones.
 $$x^2-2x+4=0\\ a=1\\ b=-2\\ c=4$$ $$x^2+2x+4=0\\ a=1\\ b=2\\ c=4$$

Neither trinomial can be factored, so we will have to rely on other methods. I'll use the quadratic formula in this case.
 $$x = {-(-2) \pm\sqrt{(-2)^2-4(1)(4)} \over 2(1)}$$ $$x = {-(2) \pm\sqrt{(2)^2-4(1)(4)} \over 2(1)}$$

Simplify both completely.
 $$x = {2 \pm\sqrt{-12} \over 2}$$ $$x = {-2 \pm\sqrt{-12} \over 2}$$

Now, convert the square root of a negative number to the simplest form in terms in i.
 $$x=\frac{2\pm 2i\sqrt{3}}{2}$$ $$x=\frac{-2\pm 2i\sqrt{3}}{2}$$

In both cases, the numerator and denominator both have common factors of two.
 $$x_{3,4}=1\pm i\sqrt{3}$$ $$x_{5,6}=-1\pm i\sqrt{3}$$

Since the original equation is written in the sixth degree, I know that the maximum number of roots equals the degree, or 6. This means I have found every root.

Let's consolidate those roots together.

$$x_{1,2}=\pm 2\\ x_{3,4}=1\pm i\sqrt{3}\\ x_{5,6}=-1\pm i\sqrt{3}$$

.
Feb 3, 2018
#4
0

find the quotion.

6x4x60x68x45+45x23+34x34x5x67x78x90x98=

Feb 5, 2018
#5
0

I do not see how this question is relevant to quotients, but I will evaluate the expression assuming that the "x" represents a multiplication operator.

6*4*60*68*45+45*23+34*34*5*67*78*90*98=266423797035

TheXSquaredFactor  Feb 6, 2018