Question

|x^2+2x+1|=4x-3

|x^2+5x|=4

Answer 1

a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{4}\\\left(x^2+2x+1-4x+3\right)\left(x^2+2x+1+4x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{4}\\\left(x^2-2x+4\right)\left(x^2+6x-2\right)=0\end{matrix}\right.\Leftrightarrow x=-3+\sqrt{11}\)

b: \(\Leftrightarrow\left[{}\begin{matrix}x^2+5x=4\\x^2+5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x-4=0\\x^2+5x+4=0\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{-1;-4;\dfrac{-5+\sqrt{41}}{2};\dfrac{-5-\sqrt{41}}{2}\right\}\)