Question

b)4x^2-1=(2x+1) (3x-5)

c)(x+1)^2= 4(x^2-2x+1)

 

Answer 1

b) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=\left(2x+1\right)\left(3x-5\right)\\ \Rightarrow2x-1=3x-5\\ \Rightarrow-x=-4\\ \Rightarrow x=4\)

c) \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)

\(\Rightarrow x^2+2x+1=4x^2-8x+4\)

\(\Rightarrow-3x^2+10x-3=0\\ \Rightarrow\left(-3x^2+9x\right)+\left(x-3\right)=0\\ \Rightarrow-3x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(1-3x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Answer 2

b, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)

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c, \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)

\(\Leftrightarrow-3x^2+10x-3=0\)

\(\Leftrightarrow-3x^2+x+9x-3=0\)

\(\Leftrightarrow x\left(-3x+1\right)-3\left(-3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

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