Step 1

To solve the equation: \((\sqrt{x^{2}-3x})=2x-6\)

Solution:

\((\sqrt{x^{2}-3x})=2x-6\)

on squaring both sides we get:

\(\Rightarrow (\sqrt{x^{2}-3x})^{2}=(2x-6)^{2}\)

\(\Rightarrow x^{2}-3x=(2x)^{2}-2(2x)(6)+6^{2}\)

\(\Rightarrow x^{2}-3x=4x^{2}-24x+36\)

\(\Rightarrow x^{2}-4x^{2}-3x+24x-36=0\)

\(\Rightarrow -3x^{2}+21x-36=0\)

\(\Rightarrow -3x^{2}+12x+9x-36=0\)

\(\Rightarrow -3x(x-4)+9(x-4)=0\)

\(\Rightarrow (x-4)(-3x+9)=0\)

\(\Rightarrow (x-4)=0, or (-3x+9)=0\)

\(\Rightarrow x=4, or 3x=9. \Rightarrow x=3\)

Step 2

Result:

Required solutions are, \(x=\{3,4\}\)