# Solve the Following Equation and Verify Your Answer: ( X + 1 X − 4 ) 2 = X + 8 X − 2 - Mathematics

Sum

$\left( \frac{x + 1}{x - 4} \right)^2 = \frac{x + 8}{x - 2}$

#### Solution

$\left( \frac{x + 1}{x - 4} \right)^2 = \frac{x + 8}{x - 2}$

$\text{ or }\frac{x^2 + 2x + 1}{x^2 - 8x + 16} = \frac{x + 8}{x - 2} [(a + b )^2 = a^2 + b^2 + 2ab\text{ and }(a - b )^2 = a^2 + b^2 - 2ab ]$

$\text{ or }x^3 + 2 x^2 + x - 2 x^2 - 4x - 2 = x^3 - 8 x^2 + 16x + 8 x^2 - 64x + 128 [\text{ After cross multiplication }]$

$\text{ or }x^3 - x^3 - 3x + 48x = 128 + 2$

$\text{ or }45x = 130$

$\text{ or }x = \frac{130}{45} = \frac{26}{9}$

$\text{ Thus }x = \frac{26}{9}\text{ is the solution of the given equation .}$

$\text{ Check: }$

$\text{ Substituting }x = \frac{26}{9} \text{ in the given equation, we get: }$

$\text{ L . H . S . }= \left( \frac{\frac{26}{9} + 1}{\frac{26}{9} - 4} \right)^2 = \left( \frac{26 + 9}{26 - 36} \right)^2 = \frac{1225}{100} = \frac{49}{4}$

$\text{ R . H . S .} = \left( \frac{\frac{26}{9} + 8}{\frac{26}{9} - 2} \right) = \left( \frac{26 + 72}{26 - 18} \right) = \frac{98}{8} = \frac{49}{4}$

$\therefore\text{ L . H . S . = R . H . S . for }x = \frac{26}{9}$

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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 9 Linear Equation in One Variable
Exercise 9.3 | Q 15 | Page 17