Question

Find the square root of \[a+x+\sqrt{2ax+{{x}^{2}}}\]

Answer 1

Hint: To find the square root of \[a+x+\sqrt{2ax+{{x}^{2}}}\]first take the \[x\] common in square root of \[2ax+{{x}^{2}}\]and then take \[2\] in both denominator and numerator to simplify the equation of square root of \[2ax+{{x}^{2}}\]to get equation \[2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}}\]and then take \[2\]in both denominator and numerator to simplify the equation of \[a+x\]to get equation \[=\dfrac{x}{2}+\dfrac{2a+x}{2}\] solving adding equations getting \[\pm {{\left( \sqrt{\dfrac{x}{2}}+\sqrt{\dfrac{2a+x}{2}} \right)}^{2}}\]by using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].

Complete step-by-step answer:
Using Algebra simplification and formula of \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[\begin{align}
  & a+x+\sqrt{2ax+{{x}^{2}}} \\
 & =a+x+\sqrt{x\left( 2a+x \right)} \\
 & =a+x+2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}} \\
 & =\dfrac{2}{2}\left( a+x \right)+2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}} \\
 & =\dfrac{2a+2x}{2}+2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}} \\
 & =\dfrac{2a}{2}+\dfrac{x}{2}+\dfrac{x}{2}+2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}} \\
 & =\dfrac{x}{2}+\dfrac{2a+x}{2}+2\sqrt{\dfrac{x}{2}\times \dfrac{\left( 2a+x \right)}{2}} \\
 & ={{\left( \sqrt{\dfrac{x}{2}}+\sqrt{\dfrac{2a+x}{2}} \right)}^{2}} \\
\end{align}\]
Using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
Therefore, the square root of \[a+x+\sqrt{2ax+{{x}^{2}}}\]
\[=\pm {{\left( \sqrt{\dfrac{x}{2}}+\sqrt{\dfrac{2a+x}{2}} \right)}^{2}}\]

Note: Simplify the equation step by step use formula according to equation and
Use formula like \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\],\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}.\]