How do you solve $2 {(x - 2)}^{\frac{2}{3}} = 50$ $2(x-2)^(2/3) = 50$ ?

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Answer 1 · 2015-04-10T12:51:35

If $2 {(x - 2)}^{\frac{2}{3}} = 50$ $2(x-2)^(2/3) = 50$
then
${(x - 2)}^{\frac{2}{3}} = 25$ $(x-2)^(2/3) = 25$

If we cube both sides we get
${(x - 2)}^{2} = 25^{3} = 25^{2} \cdot 5^{2}$ $(x-2)^2 = 25^3 = 25^2*5^2$

or
${(x - 2)}^{2} - 125^{2} = 0$ $(x-2)^2 - 125^2 = 0$

remember that anything of the form
$(a^{2} - b^{2})$ $(a^2-b^2)$ can be factored as $(a - b) (a + b)$ $(a-b)(a+b)$

we have
$((x - 2) - 125) ((x - 2) + 125) = 0$ $((x-2)-125)((x-2)+125) = 0$
or
$(x - 127) (x + 123) = 0$ $(x-127)(x+123)=0$

so
$x = 127$ $x=127$
or
$x = - 123$ $x=-123$

How do you solve 2(x−2)23=502(x-2)^(2/3) = 50?