Question

Solve for x: $\sqrt {ax + 1} - \sqrt {ax - 1} = \sqrt {ax} $
A. $\dfrac{{ - 2}}{{3a}}$
B. $\dfrac{2}{{3a}}$
C. $\dfrac{{ \pm 2\sqrt 3 }}{{3a}}$
D. $\dfrac{{2\sqrt 3 }}{{3a}}$

Answer 1

Hint: In this question, first we will square on both sides so that the square root gets removed. After this rearrange the terms on both sides and we get remaining square root terms on one side and without square root terms on the other side. Again take square on both sides and solve further to get the value of x.

Complete step-by-step answer:

The given equation is:

$\sqrt {ax + 1} - \sqrt {ax - 1} = \sqrt {ax} $

On squaring both sides, we get:

${\text{ax + 1 + ax - 1 - 2(}}\sqrt {{\text{ax + 1}}} )(\sqrt {{\text{ax - 1}}} ) = {\text{ax}}$

Taking the square root terms to RHS, we get:

$

  {\text{2ax - ax = 2(}}\sqrt {{\text{ax + 1}}} )(\sqrt {{\text{ax - 1}}} ) \\

   \Rightarrow {\text{ax = 2(}}\sqrt {{\text{ax + 1}}} )(\sqrt {{\text{ax - 1}}} ) \\

 $

We know that (a+b)(a-b) = ${{\text{a}}^2} - {{\text{b}}^2}$. Therefore, we can write the above equation as:

${\text{ax = 2(}}\sqrt {{\text{ax + 1}}} )(\sqrt {{\text{ax - 1}}} ) = \sqrt {({\text{ax + 1)(ax - 1)}}}

= \sqrt {{{{\text{(ax)}}}^2} - {1^2}} = \sqrt {{{\text{a}}^2}{{\text{x}}^2} - 1} $


Again, on squaring on both sides, we get:

\[

  {({\text{ax)}}^2} = {\left( {2\sqrt {{{\text{a}}^2}{{\text{x}}^2} - 1} } \right)^2} = 4\left(

{{{\text{a}}^2}{{\text{x}}^2} - 1} \right) = 4{{\text{a}}^2}{{\text{x}}^2} - 4 \\

   \Rightarrow 4{{\text{a}}^2}{{\text{x}}^2} - {{\text{a}}^2}{{\text{x}}^2} = 4. \\

   \Rightarrow 3{{\text{a}}^2}{{\text{x}}^2} = 4 \\

   \Rightarrow {{\text{x}}^2} = \dfrac{4}{{3{{\text{a}}^2}}} \\

   \Rightarrow {\text{x = }} \pm \sqrt {\dfrac{4}{{3{{\text{a}}^2}}}} = \pm

\dfrac{2}{{{\text{a}}\sqrt 3 }} \\

\]

On rationalising the above value of ‘x’, we get:

x = \[ \pm \dfrac{2}{{{\text{a}}\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \pm \dfrac{{2\sqrt 3 }}{{3a}}\]

But a negative value of ‘x’ is not permissible as it results in complex values.

$\therefore $ x =\[\dfrac{{2\sqrt 3 }}{{3a}}\]

Therefore, the option D is correct.

Note: In the question involving square roots on both sides, the most important step is to square on both sides of the equation. In this question to further simplify, you should remember the algebraic identities like ${\left( {{\text{a + b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab and }}\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right) = {{\text{a}}^2} - {{\text{b}}^2}$.You must check the values of ‘x’ by putting in original equation.