Question

Solve for $x:\dfrac{\sqrt{36x+1}+6\sqrt{x}}{\sqrt{36x+1}-6\sqrt{x}}=9$

Answer 1

Hint: Rationalize the given expression, apply basic identities of ${{\left( a+b \right)}^{2}}$ and ${{a}^{2}}-{{b}^{2}}$ and simplify the expression and get the value of x.

Complete step-by-step answer:
Rationalize it by taking the conjugate of denominator and multiply it to the expression on both numerator and denominator.
We have been given an expression which we need to solve and get the value of x.
$\dfrac{\sqrt{36x+1}+6\sqrt{x}}{\sqrt{36x+1}-6\sqrt{x}}=9$
Let us first rationalize the above expression i.e. multiply the numerator and denominator with $\left( \sqrt{36x+1}+6\sqrt{x} \right)$ in the LHS of the expression.
\[=\left[ \dfrac{\sqrt{36x+1}+6\sqrt{x}}{\sqrt{36x+1}-6\sqrt{x}} \right]\times \left[ \dfrac{\sqrt{36x+1}+6\sqrt{x}}{\sqrt{36x+1}-6\sqrt{x}} \right]\]
Let us simplify the expression using the basic identities,
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
Here \[a=\sqrt{36x+1}\] and \[b=6\sqrt{x}\]
Thus, the expression forms
LHS \[=\dfrac{{{\left( \sqrt{36x+1}+6\sqrt{x} \right)}^{2}}}{\left( \sqrt{36x+1}-6\sqrt{x} \right)\left( \sqrt{36x+1}+6\sqrt{x} \right)}\]
\[=\dfrac{\left( 36x+1 \right)+\left( 36x \right)+2\left( \sqrt{36x+1} \right)\left( 6\sqrt{x} \right)}{{{\left( \sqrt{36x+1} \right)}^{2}}-{{\left( 6\sqrt{x} \right)}^{2}}}\]
\[=\dfrac{72x+1+12\sqrt{x}\sqrt{36x+1}}{36x+1-36x}=\dfrac{72x+1+12\sqrt{36{{x}^{2}}+x}}{1}\]
\[\Rightarrow 72x+12\sqrt{36{{x}^{2}}+x}+1=9\]

Thus, let us simplify it as
\[72x+12\sqrt{36{{x}^{2}}+x}=9-1\]
\[72x+12\sqrt{36{{x}^{2}}+x}=8\]
\[12\left[ 6x+\sqrt{36{{x}^{2}}+x} \right]=8\]
\[6x+\sqrt{36{{x}^{2}}+x}=\dfrac{8}{12}=\dfrac{2}{3}\]
\[6x+\sqrt{36{{x}^{2}}+x}=\dfrac{2}{3}\Rightarrow \sqrt{36{{x}^{2}}+x}=\dfrac{2}{3}-6x\]
Let us square both sides of the above equation,
\[{{\left( \sqrt{36{{x}^{2}}+x} \right)}^{2}}={{\left( \dfrac{2}{3}-6x \right)}^{2}}\]
\[36{{x}^{2}}+x=\dfrac{4}{9}-2\times \dfrac{2}{3}\times 6x+36{{x}^{2}}\]
Cancel out \[36{{x}^{2}}\] from LHS and RHS and simplify the expression
\[+x=\dfrac{4}{9}-8x\]
\[\Rightarrow 8x+x=\dfrac{4}{9}\]
\[9x=\dfrac{4}{9}\]
\[x=\dfrac{4}{9\times 9}=\dfrac{4}{81}\]
Thus, we got the required value as \[x=\dfrac{4}{81}\] .
Hence, \[x=\dfrac{4}{81}\] is the solution for the given expression.

Note: Remember that for an expression like this it's important to rationalize. If any expression of root comes in the denominator, then rationalize the entire expression by the conjugate of the denominator.