Question

I need help. this is the problem x^2/3+10=7x^1/3

Answer 1

$\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow (x^{\frac{1}{3}})^2-7x^{\frac{1} {3}} +10=0 \\\;\\ We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be: \\\;\\ t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0$

$\it \Leftrightarrow (t-2)(t-5)=0 \\\;\\ t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8 \\\;\\ t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5 \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125$

S = {8; 125}