What is $x$ $x$ in ${(x - 2)}^{2} = 64$ $(x-2)^2=64$ ?

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Answer 1 · 2018-05-24T23:04:26

$x = - 6 or x = 10$ $x=-6 or x=10$

${(x - 2)}^{2} = 64$ $(x - 2)^2 = 64$

The exponent $2$ $2$ means that $x - 2$ $x-2$ will multiply itself twice.

$(x - 2) (x - 2) = 64$ $(x - 2)(x - 2) = 64$

Use the distributive property on the left side

$(x) (x) + (x) (- 2) + (- 2) (x) + (- 2) (- 2) = 64$ $(x)(x) + (x)(-2) + (-2)(x) + (-2)(-2)=64$

$x^{2} - 2 x - 2 x + 4 = 64$ $x^2 - 2x - 2x +4 = 64$

$x^{2} - 4 x + 4 = 64$ $x^2 - 4x + 4 = 64$

Now we can subtract $64$ $64$ from both sides

$x^{2} - 4 x + 4 - 64 = 64 - 64$ $x^2 - 4x + 4 - 64 = 64 - 64$

$x^{2} - 4 x - 60 = 0$ $x^2 - 4x - 60 = 0$

Then factorize the left side

$(x + 6) (x - 10) = 0$ $(x + 6)(x-10)=0$

Now we can set the factors equal to $0$ $0$

$x + 6 = 0 or x - 10 = 0$ $x + 6 = 0 or x - 10 = 0$

$x = 0 - 6 or x = 0 + 10$ $x = 0 - 6 or x = 0 + 10$

$x = - 6 or x = 10$ $x = -6 or x = 10$ $\leftarrow$ $larr$ This is the final answer!

What is xxx in (x-2)^2=64(x−2)2=64(x-2)^2=64 ?