# 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

Answer

Verified

384.3k+ views

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.

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.

Recently Updated Pages

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

Which country launched the first satellite in space class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE