Question

find the value of x:
$1. \sqrt[3]{2x - 1} - 3 = 0 \\ 2. \sqrt[5]{3x + 2} - 2 = 0$
​

Answer 1

$\large\underline{\sf{Solution-i}}$

$\rm :\longmapsto\: \sqrt[3]{2x - 1} - 3 = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt[3]{2x - 1} = 3$

On Cubing both sides, we get

$\rm :\longmapsto\:2x - 1 = {(3)}^{3}$

$\rm :\longmapsto\:2x - 1 =27$

$\rm :\longmapsto\:2x =27 + 1$

$\rm :\longmapsto\:2x =28$

$\bf\implies \:x = 14$

Verification :-

Consider LHS

$\rm :\longmapsto\: \sqrt[3]{2x - 1} - 3$

On Substituting x = 14, we get

$\rm \: = \: \: \sqrt[3]{2 \times 14 - 1} - 3$

$\rm \: = \: \: \sqrt[3]{28 - 1} - 3$

$\rm \: = \: \: \sqrt[3]{27} - 3$

$\rm \: = \: \: \sqrt[3]{3 \times 3 \times 3} - 3$

$\rm \: = \: \: 3 - 3$

$\rm \: = \: \:0$

Hence, Verified

$\red{\large\underline{\sf{Solution-ii}}}$

$\rm :\longmapsto\: \sqrt[5]{3x + 2} - 2 = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt[5]{3x + 2} = 2$

On taking 5 power on both sides, we get

$\rm :\longmapsto\:3x + 2 = {(2)}^{5}$

$\rm :\longmapsto\:3x + 2 = 32$

$\rm :\longmapsto\:3x = 32 - 2$

$\rm :\longmapsto\:3x = 30$

$\bf\implies \:x = 10$

Verification :-

Consider LHS

$\rm :\longmapsto\: \sqrt[5]{3x + 2} - 2$

On substituting x = 10, we get

$\rm \: = \: \: \sqrt[5]{3 \times 10 + 2} - 2$

$\rm \: = \: \: \sqrt[5]{30 + 2} - 2$

$\rm \: = \: \: \sqrt[5]{32} - 2$

$\rm \: = \: \: \sqrt[5]{2 \times 2 \times 2 \times 2 \times 2} - 2$

$\rm \: = \: \: 2 - 2$

$\rm \: = \: \:0$

Hence, Verified