How do you solve then check for extraneous solutions: ${(x + 5)}^{\frac{2}{3}} = 4$ $(x+5)^(2/3)=4$ ?

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Answer 1 · 2018-03-25T13:19:36

$x = 3$ $x=3$

Raise both sides to the power of $\frac{3}{2}$ $3/2$

${({(x + 5)}^{\frac{2}{3}})}^{\frac{3}{2}} = 4^{\frac{3}{2}}$ $((x+5)^(2/3))^(3/2) = 4^(3/2)$

$x + 5 = {(\pm \sqrt{4})}^{3}$ $x+5 = (+-sqrt4)^3$

There are two possible solutions for $\sqrt{4}$ $sqrt4$

$x + 5 = {(+ 2)}^{3} or x + 5 = {(- 2)}^{3}$ $x+5 = (+2)^3" "or" "x+5 = (-2)^3$

$x + 5 = 8 or x + 5 = - 8$ $" "x+5 = 8" "or" "x+5 = -8$

$x = 8 - 5 or x = - 8 - 5$ $x = 8-5" "or" "x = -8-5$

$x = 3 or x = - 13$ $x = 3 or x=-13$

Check both in the original equation.

${(3 + 5)}^{\frac{2}{3}} = 8^{\frac{2}{3}} = 4$ $(3+5)^(2/3) = 8^(2/3) = 4$

${(13 + 5)}^{\frac{2}{3}} = 13^{\frac{2}{3}} = 1.768 \leftarrow$ $(13+5)^(2/3) = 13^(2/3) =1.768" "larr$ this does not give $4$ $4$

How do you solve then check for extraneous solutions: (x+5)^(2/3)=4(x+5)23=4(x+5)^(2/3)=4?