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# Help me asap

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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

#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
+100483
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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
+2338
+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.

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

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Feb 3, 2018
#4
0

find the quotion.

6x4x60x68x45+45x23+34x34x5x67x78x90x98=

Feb 5, 2018
#5
+2338
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