2x5/3 +64=0

Question

2x^5/3+64=0

Answer 1

     2x^5/3 + 64 = 0

Subtract 64 from both sides of the equation:

     2x^5/3 = -64

Divide both sides of the equation by 2:

     x^5/3 = -32

Cube both sides of the equation:

    (x^5/3)³ = (-32)³

          x⁵ = -32768

Take both sides of the equation to the power of 1/5:

     (x⁵)^1/5 = (-32768)^1/5

             x = -8

Answer 2

This problem may have been written incorrectly, as it only has complex number solutions (not part of algebra 1).

The exact (complex) solutions are:

x = 8(-1)^3/5

x = -8(-1)^2/5

 

Note: if you change the plus sign in 2x^5/3 + 64=0 to a minus sign:

 2x^5/3 - 64=0

then follow Tamara's working, you'll get x = 8

Answer 3

2x^(5/3)+64=0 Solution 2x5/3+64=0 x^(5/3) = -64/2 x^(5/3) = -32 x^5 = -32^3 x^5 = -32768 x = (-32768)^(1/5) x = (32768)^(1/5) i, where i = (-1)^(1/2) x=8i =========================================================================== Step by Step solution Our goald is to write an equation with a form x = ? 1st step, sume -64 both places of the equation. That is because you do not want the +64 at the left side of the equation. 2x^(5/3) = -64 2nd step, divide by 1/2 all the equation. That is because you do not want 2x, you want x only x^(5/3) = -64/2 = -32 x^(5/3) = -64/2, then x^(5/3) = -32 3rd step, raises the whole equation to 3/5 exponent [x^(5/3) = -32]^(3/5) [x^(5/3)]^(3/5) = [-34]^(3/5) 4st stepUse the exponent property (a^n)^m = a^(nm) x^(5*3/3*5)= [-34]^(3/5) x = [-34]^(3/5)= R(5) [-34^3], wher R()[?]= fifth root of ? note: R(5)[34^3*(-1)^3]=R(5)[39304*-1]=R(5)[39304]i x=8i