Question

Find the value of x in each of the following:
(i) $\sqrt[3]{34x-7}-5=0$ (ii) $\sqrt[4]{43x+1}=2$

Answer 1

(i) $\sqrt[3]{34x-7}-5=0$

$\Rightarrow \sqrt[3]{34x-7}=5$

$\Rightarrow (\sqrt[3]{34x-7}{)}^3=5^3$

$\Rightarrow 4x-7=125$ $[\because (\sqrt[n]{na}{)}^n=a]$

$\Rightarrow 4x=132$

$\Rightarrow x=33$

(ii) $\sqrt[4]{43x+1}=2$

$\Rightarrow (\sqrt[4]{43x+1}{)}^4=2^4$

$\Rightarrow 3x+1=16$ $[\because (\sqrt[n]{na}{)}^n=a]$

$\Rightarrow 3x=15$

$x=5$