solve x^6 - 64 = 0
(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)
x ^ 6 - 64 = 0 Add 64 to both sides
x ^ 6 - 64 + 64 = 0 + 64
x ^ 6 = 64
x = sixth root of 64
x = + OR - 2
Solutions :
x = - 2
and
x = 2
since you have a x^6 .... there will be 6 solutions including complex numbers.
Bosnian gave you the two real roots,
Monic only factored the expression, using the factored form we can find the 4 complex roots
x^2 - 2x + 4 = 0
x^2 - 2x + 1 = -4+1
(x-1)^2 = -3
x-1 = ±√3 i
x = 1 ± √3 i
In a similar way
x^2 + 2x + 4 = 0 gives us
x = -1 ±√3 i
(when solving quadratics with a coefficient of 1 for the x^2 term and an EVEN middle coefficient, I always use completing the square rather than the formula, it is faster and yields the answer in reduced form)
x+8<4 help me self this please,
To solve the equation x^6 - 64 = 0, we can use the concept of factoring the difference of squares.
Step 1: Recognize that 64 can be expressed as a perfect square, which is 8^2.
Step 2: Rewrite the equation as (x^3)^2 - 8^2 = 0.
Step 3: Apply the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b). In this case, a = x^3 and b = 8.
So, (x^3 + 8)(x^3 - 8) = 0.
Step 4: Set each factor equal to zero and solve for x:
x^3 + 8 = 0 or x^3 - 8 = 0.
For the first factor, x^3 + 8 = 0, subtract 8 from both sides to get x^3 = -8.
Then, taking the cube root of both sides, we find x = -2.
For the second factor, x^3 - 8 = 0, add 8 to both sides to get x^3 = 8.
Again, taking the cube root of both sides, we find x = 2.
So, the solutions to the equation x^6 - 64 = 0 are x = -2 and x = 2.
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