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# Solve the following quadratic equation by factorization:$\dfrac{{5 + x}}{{5 - x}} - \dfrac{{5 - x}}{{5 + x}} = 3\dfrac{3}{4};x \ne 5, - 5$The roots are $3,\dfrac{{ - 25}}{3}$.(a) True (b) False

Last updated date: 04th Aug 2024
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Hint: Rationalize the identity so that the evaluation becomes simple. Also, the standard identities can be easily used.

We have the given quadratic equation as:
$\dfrac{{5 + x}}{{5 - x}} - \dfrac{{5 - x}}{{5 + x}} = 3\dfrac{3}{4}$
By rationalization, we get,
$\Rightarrow \dfrac{{5 + x}}{{5 - x}}\left( {\dfrac{{5 + x}}{{5 + x}}} \right) - \dfrac{{5 - x}}{{5 + x}}\left( {\dfrac{{5 - x}}{{5 - x}}} \right) = \dfrac{{15}}{4}$
$\Rightarrow \dfrac{{{{(5 + x)}^2} - {{(5 - x)}^2}}}{{(5 + x)(5 - x)}} = \dfrac{{15}}{4}$ … (1)
Now, we know the identities,
${(a + b)^2} = {a^2} + {b^2} + 2ab$
${(a - b)^2} = {a^2} + {b^2} - 2ab$
$({a^2} - {b^2}) = (a - b)(a + b)$
Therefore, by using these identities, we get the equation (1) as
$\Rightarrow \dfrac{{({{(5)}^2} + 10x + {x^2}) - ({{(5)}^2} - 10x + {x^2})}}{{({{(5)}^2} - {{(x)}^2})}} = \dfrac{{15}}{4}$
$\Rightarrow \dfrac{{(25 + 10x + {x^2} - 25 + 10x - {x^2})}}{{(25 - {x^2})}} = \dfrac{{15}}{4}$
$\Rightarrow \dfrac{{20x}}{{(25 - {x^2})}} = \dfrac{{15}}{4}$
Multiplying both sides by $\dfrac{1}{5}$ , we get,
$\Rightarrow \dfrac{{20x}}{{(25 - {x^2})}} \times \left( {\dfrac{1}{5}} \right) = \dfrac{{15}}{4} \times \left( {\dfrac{1}{5}} \right)$
$\Rightarrow \dfrac{{4x}}{{(25 - {x^2})}} = \dfrac{3}{4}$
$\Rightarrow (4x) \times 4 = 3 \times (25 - {x^2})$
$\Rightarrow 16x = 75 - 3{x^2}$
$\Rightarrow 3{x^2} + 16x - 75 = 0$
$\Rightarrow 3{x^2} + 25x - 9x - 75 = 0$
$\Rightarrow x(3x + 25) - 3(3x + 25) = 0$
$\Rightarrow (x - 3)(3x + 25) = 0$
Now, either $(x - 3) = 0$
$\therefore x = 3$
Or $(3x + 25) = 0$
$\therefore x = \dfrac{{ - 25}}{3}$
Hence, the roots are $3,\dfrac{{ - 25}}{3}$.
So, the required solution is (a) True.

Note: In order to solve these types of questions, an adequate knowledge of standard identities is needed, after substituting these identities in the quadratic equations, further evaluation will lead to the desired result.