Solve the equation 8 x - 5 3 x - 7 = 3 x - 7 x - 5 .

Given 8 x - 5 3 x - 7 = 3 x - 7 x - 5 Here, x ≠ 7 3 , 5 Cross multiplying we get 8 ( x - 5 ) 3 2 = ( 3 x - 7 ) 3 2 Raising both sides to the power 2 3 we get 8 2 3 ( x - 5 ) = ( 3 x - 7 ) ⇒ 4 ( x - 5 ) = ( 3 x - 7 ) ⇒ x = 13

Solve the equation 18 ( 7 x - 3 ) 2 x + 1 = 250 2 x + 1 3 7 x - 3 .

Given 18 ( 7 x - 3 ) 2 x + 1 = 250 2 x + 1 3 7 x - 3 Cross multiplying, we get 54 ( 7 x - 3 ) 3 2 = 250 2 x + 1 3 2 ⇒ 27 ( 7 x - 3 ) 3 2 = 125 ( 2 x + 1 ) 3 2 ⇒ ( 3 ) 3 ( 7 x - 3 ) 3 2 = ( 5 ) 3 ( 2 x + 1 ) 3 2 Raising both sides to the power 2 3 , we get 9 ( 7 x - 3 ) = 25 ( x + 1 ) ⇒ 13 x = 52 ∴ x = 4 .

Solve the equation ( a + x ) 2 3 + 4 ( a - x ) 2 3 = 5 a 2 - x 2 1 3 .

Given ( a + x ) 2 3 + 4 ( a - x ) 2 3 = 5 a 2 - x 2 1 3 ⇒ ( a + x ) 2 / 3 a 2 - x 2 1 / 3 + 4 ( a - x ) 2 / 3 a 2 - x 2 1 / 3 = 5 a 2 - x 2 1 / 3 a 2 - x 2 1 / 3 ⇒ ( a + x ) 1 / 3 ( a - x ) 1 / 3 + 4 · ( a - x ) 1 / 3 ( a + x ) 1 / 3 = 5 Putt a + x a - x 1 / 3 = p then, we have p + 4 p = 5 ⇒ p 2 - 5 p + 4 = 0 ⇒ ( p - 4 ) ( p - 1 ) = 0 ⇒ p = 4 , 1 Hence if a + x a - x 1 / 3 = p = 1 ⇒ a + x a - x = 1 ⇒ a + x = a - x ∴ x = 0 Again, if a + x a - x 1 / 3 = p = 4 ⇒ a + x a - x = 64 ⇒ 65 x = 63 a ∴ x = 63 65 a Hence, the roots of the given equation are x = 0 , 63 a 65 .

Solve the equation x 2 + a x - 1 - x 2 + b x - 1 = a - b .

x 2 + a x - 1 - x 2 + b x - 1 = a - b . . . ( i ) Also, we can write as x 2 + a x - 1 - x 2 + b x - 1 = x ( a - b ) . . . ii Factorizing it, we get x 2 + a x - 1 - x 2 + b x - 1 × x 2 + a x - 1 + x 2 + b x - 1 = x ( a - b ) ⇒ a - b x 2 + a x - 1 + x 2 + b x - 1 = x a - b a + b ⇒ x 2 + a x - 1 + x 2 + b x - 1 = x a + b . . . . iii Adding ( i ) and ii i , we get 2 x 2 - a x - 1 = a ( x + 1 ) + b ( x - 1 ) Squaring both sides, we have 4 x 2 + a x - 1 = a x + 1 2 + b x - 1 2 + 2 a b x 2 - 1 ⇒ 4 x 2 + a x - 1 = a x 2 + 2 x + 1 + b x 2 - 2 x + 1 + 2 a b x 2 - 1 ⇒ 4 x 2 + a x - 1 = a + b + 2 a b x 2 + 2 a - 2 b x + a + b - 2 a b Taking terms to one side, we have x 2 ( a + b + 2 a b - 4 ) + 2 x a - b - 2 a + a + b - 2 a b + 4 = 0 ⇒ x 2 ( a + b ) 2 - 4 - 2 x a + b + a - b 2 + 4 = 0 Solving this for x , we have x = 2 ( a + b ) ± 4 ( a + b ) 2 - 4 a + b 2 - 4 × a - b 2 + 4 2 a + b 2 - 4 On further simplification, we get x = 1 , a - b 2 + 4 a + b 2 - 4

Solve the equation x + x 2 - 1 x - x 2 - 1 + x - x 2 - 1 x + x 2 - 1 = 98 .

Given x + x 2 - 1 x - x 2 - 1 + x - x 2 - 1 x + x 2 - 1 = 98 ⇒ x + x 2 - 1 2 + x - x 2 - 1 2 x 2 - x 2 - 1 = 98 ⇒ x 2 + x 2 - 1 + 2 x x 2 - 1 + x 2 + x 2 - 1 - 2 x x 2 - 1 1 = 98 ⇒ 4 x 2 - 2 = 98 ⇒ 4 x 2 = 100 ⇒ x 2 = 25 ∴ x = ± 5

E M B I B E

Higher Algebra>Chapter 10 - Miscellaneous Equations>EXAMPLES. X.a.

H S Hall and S R Knight Solutions for Chapter: Miscellaneous Equations, Exercise 1: EXAMPLES. X.a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Miscellaneous Equations, Exercise 1: EXAMPLES. X.a.

Attempt the free practice questions on Chapter 10: Miscellaneous Equations, Exercise 1: EXAMPLES. X.a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.