Question

# Solve the following equation: $\sqrt {\dfrac{{\text{x}}}{{{\text{1 - x}}}}} {\text{ + }}\sqrt {\dfrac{{{\text{1 - x}}}}{{\text{x}}}} {\text{ = 2}}\dfrac{1}{6}$

Hint: Taking L.C.M. of R.H.S. , will give $({\text{x + 1 - x)}}$ term in numerator.
Given, $\sqrt {\dfrac{{\text{x}}}{{{\text{1 - x}}}}} {\text{ + }}\sqrt {\dfrac{{{\text{1 - x}}}}{{\text{x}}}} {\text{ = 2}}\dfrac{1}{6}$ -- (1)

We need to find the value of x by solving the equation (1). For solving the equation we are going to use the law of surds which states $\sqrt {\dfrac{{\text{a}}}{{\text{b}}}} {\text{ = }}\dfrac{{\sqrt {\text{a}} }}{{\sqrt {\text{b}} }}$ .On applying this law on equation (1) we get, $\dfrac{{\sqrt {\text{x}} }}{{\sqrt {{\text{1 - x}}} }}{\text{ + }}\dfrac{{\sqrt {{\text{1 - x}}} }}{{\sqrt {\text{x}} }}{\text{ = 2}}\dfrac{1}{6}$
Now, for further simplification we are taking L.C.M. on the left hand side.
$\dfrac{{\sqrt {\text{x}} \times \sqrt {\text{x}} {\text{ + }}\sqrt {1{\text{ - x}}} \times \sqrt {{\text{1 - x}}} }}{{\sqrt {{\text{1 - x}}} \times \sqrt {\text{x}} }}{\text{ = 2}}\dfrac{1}{6}$ -- (2)

Another law of surds states that $\sqrt {\text{a}} \times \sqrt {\text{b}} {\text{ = }}\sqrt {{\text{a}} \times {\text{b}}}$ . Using this law on equation (2), we get
$\dfrac{{\sqrt {{\text{x}} \times {\text{x}}} {\text{ + }}\sqrt {\left( {{\text{1 - x}}} \right) \times \left( {{\text{1 - x}}} \right)} }}{{\sqrt {\left( {{\text{1 - x}}} \right) \times {\text{x}}} }}{\text{ = 2}}\dfrac{1}{6}$ . And from law of indices we know that ${{\text{a}}^{\text{m}}}{\text{ }} \times {\text{ }}{{\text{a}}^{\text{n}}}{\text{ = }}{\left( {\text{a}} \right)^{{\text{m + n}}}}$.
In our case m = n = 1. Hence, we get
$\dfrac{{\sqrt {{{\text{x}}^2}} {\text{ + }}\sqrt {{{\left( {{\text{1 - x}}} \right)}^2}} }}{{\sqrt {\left( {1{\text{ - x}}} \right) \times {\text{x}}} }}{\text{ = 2}}\dfrac{1}{6}$
Now, we know that $\sqrt {\text{a}} {\text{ = }}{{\text{a}}^{\dfrac{1}{2}}}$ and another law of indices states that ${\left( {{{\text{a}}^{\text{m}}}} \right)^{\text{n}}}{\text{ = }}{{\text{a}}^{{\text{mn}}}}$. Applying these laws on above equation, we get
$\dfrac{{{{\text{x}}^{2 \times \dfrac{1}{2}}}{\text{ + }}{{\left( {1{\text{ - x}}} \right)}^{2 \times \dfrac{1}{2}}}}}{{{{\left( {{\text{1 - x}}} \right)}^{\dfrac{1}{2}}} \times {{\text{x}}^{\dfrac{1}{2}}}}}{\text{ = 2}}\dfrac{1}{6}$ . Solving it further, we get

$\dfrac{{{\text{x + 1 - x}}}}{{{{\left( {{\text{1 - x}}} \right)}^{\dfrac{1}{2}}} \times {{\text{x}}^{\dfrac{1}{2}}}}}{\text{ = 2}}\dfrac{1}{6}$
$\dfrac{1}{{{{\left( {1{\text{ - x}}} \right)}^{\dfrac{1}{2}}} \times {{\text{x}}^{\dfrac{1}{2}}}}}{\text{ = 2}}\dfrac{1}{6}$
Since ${\text{a}}\dfrac{{\text{b}}}{{\text{c}}}{\text{ = }}\dfrac{{{\text{ac + b}}}}{{\text{c}}}$ . Applying on above equation, we get
$\dfrac{1}{{{{\left( {1{\text{ - x}}} \right)}^{\dfrac{1}{2}}} \times {{\text{x}}^{\dfrac{1}{2}}}}}{\text{ = }}\dfrac{{2 \times 6{\text{ + }}1}}{6}$
$\dfrac{1}{{{{\left( {1{\text{ - x}}} \right)}^{\dfrac{1}{2}}} \times {{\text{x}}^{\dfrac{1}{2}}}}}{\text{ = }}\dfrac{{13}}{6}$
Now, for finding x, we need to perform squaring on both the sides of the equations. On squaring we get,
$\dfrac{1}{{\left( {{\text{1 - x}}} \right) \times \left( {\text{x}} \right)}}{\text{ = }}\dfrac{{169}}{{36}}$
By cross-multiplication, we get
$169(1 - {\text{x)(x) = 36}}$
Solving further we get,
$169({\text{x - }}{{\text{x}}^2}){\text{ = 36}} \\ {\text{169x - 169}}{{\text{x}}^2}{\text{ = 36}} \\ \\$
Sending each terms on one side, we get
${\text{169}}{{\text{x}}^2}{\text{ - 169x + 36 = 0}}$ -- (3)
We have a quadratic equation in x. For finding the value of x, we are required to find the real roots of the equation. For any finding the root of any quadratic equation ${\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}$ , we can use the quadratic formula.
Quadratic formula, ${\text{ x = }}\dfrac{{ - {\text{b}} \pm \sqrt {{{\text{b}}^2} - 4{\text{ac}}} }}{{2{\text{a}}}}$ . In this formula we call the term $\sqrt {{{\text{b}}^2} - 4{\text{ac}}}$ the determinant. If the determinant is positive then only, we will have real roots of the quadratic equation.
Comparing equation (3) with ${\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}$ , we get
a = 169, b = -169 and c = 36. Applying the quadratic formula we get,
${\text{x = }}\dfrac{{169 \pm \sqrt {{{169}^2} - 4(169)(36)} }}{{2(169)}} \\ {\text{x = }}\dfrac{{169 \pm \sqrt {28561 - 24336} }}{{338}} \\ {\text{x = }}\dfrac{{169 \pm \sqrt {4225} }}{{338}} \\ {\text{x = }}\dfrac{{169 \pm 65}}{{338}} \\ {\text{x = }}\dfrac{{13 \pm 5}}{{26}} \\ {\text{x = }}\dfrac{{13 + 5}}{{26}}{\text{ and x = }}\dfrac{{13{\text{ - 5}}}}{{26}} \\ {\text{x = }}\dfrac{{18}}{{26}}{\text{ and x = }}\dfrac{8}{{26}} \\ {\text{x = }}\dfrac{9}{{13}}{\text{ and x = }}\dfrac{4}{{13}} \\$
We have two real roots of x. Now, for knowing the correct solution of the equation we need to have$\sqrt {\dfrac{{\text{x}}}{{1{\text{ - x}}}}} {\text{ > 0}}$.
For this to be true, x should lie in between (0, 1). And for both the value of x, the condition is true.
Hence, the solution of the equation is x = $\dfrac{9}{{13}}$ and $\dfrac{4}{{13}}$

Note:- In these types of equations , it is required to eliminate square roots. But if we shouldnâ€™t perform squaring in the first step .It will only complicate the problem. We should start with simplifying by using laws of indices and surds.