Question

Help!! ASAP, please!

Fill in the missing blanks for the radical equation.

Answer 1

Answer:

x = 18

Step-by-step explanation:

Given $3 \sqrt[4]{(x-2)^3}-4 = 20$

now adding '4' on both sides, we get

$3 \sqrt[4]{(x-2)^3}-4+4= 20+4$

$3 \sqrt[4]{(x-2)^3} = 24$

i)  $3 \sqrt[4]{(x-2)^3}$  = 24

cancelling '3' on both sides, we get

$\sqrt[4]{(x-2)^3} = 8$

ii)  $\sqrt[4]{(x-2)^3}$    =   8

iii) $(x-2)^{\frac{3}{4} } = 8$

$(x-2)^{\frac{3}{4} } = 2^{3}$

iv) $(x-2)^\frac{3}{4} )^{\frac{1}{3} } = (2^{3})^\frac{1}{3}$

now simplify on both sides,we get

$(x-2)^\frac{1}{4} = 2$

$((x-2)^{\frac{1}{4} })^4 = 2^4$

simplify we get solution is

v) (x-2) = 16

vi) final answer is x = 16 +2 =18