Question

GPT:

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

Answer 1

\(\frac{-1}{3}\le x\le6\\ \sqrt[]{3x+1}-4-\left(\sqrt[]{6-x}-1\right)+3x^2-14x-5=0\\ \Leftrightarrow\frac{3x-15}{\sqrt[]{3x+1}+4}+\frac{x-5}{\sqrt[]{6-x+1}}+\left(x-5\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(x-5\right)\left(\frac{3}{\sqrt[]{3x+1}}+\frac{1}{\sqrt[]{6-x}+1}+3x-1\right)=0\)

do\(x\ge\frac{-1}{3}\Rightarrow3x+1\ge0\\ \frac{3}{\sqrt[]{3x+1}}+\frac{1}{\sqrt[]{6-x}+1}+3x-1>0\\ \Rightarrow x=5\)