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

Last updated date: 23rd Jul 2024
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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 shouldn’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.