Question

a) x³- \(\dfrac{1}{4}\)x = 0

b) (2x - 1)²- (x + 3)² = 0

c) x²(x - 3) + 12 - 4x = 0

Answer 1

\(a,\Rightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Rightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\\ \Rightarrow\left(x-4\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-4\right)\left(x-3\right)=0\\ \Rightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)

Answer 2

a: \(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)