Solve equation for x by factorisation x + 1 x = 2 1 20 , x ≠ 0

Given equation: x + 1 x = 2 1 20 Convert the given equation in the form of a x 2 + b x + c = 0 ⇒ x 2 + 1 x = 41 20 ⇒ 20 x 2 + 20 = 41 x ⇒ 20 x 2 - 41 x + 20 = 0 By splitting the middle term. ⇒ 20 x 2 - 16 x - 25 x + 20 = 0 ⇒ 4 x ( 5 x - 4 ) - 5 ( 5 x - 4 ) = 0 ⇒ ( 5 x - 4 ) ( 4 x - 5 ) = 0 Now, ( 5 x - 4 ) = 0 ⇒ 5 x = 4 ⇒ x = 4 5 ( 4 x - 5 ) = 0 ⇒ 4 x = 5 ⇒ x = 5 4 Hence, x = 4 5 and x = 5 4 are the roots of the given equation.

Solve equation for x by factorisation 3 x - 8 x = 2 , x ≠ 0

Given equation: 3 x - 8 x = 2 ⇒ 3 x 2 - 8 x = 2 ⇒ 3 x 2 - 2 x - 8 = 0 By the method of splitting middle term we can write, ⇒ 3 x 2 - 6 x + 4 x - 8 = 0 ⇒ 3 x x - 2 + 4 x - 2 = 0 ⇒ x - 2 3 x + 4 = 0 ⇒ x = 2 or x = - 4 3 Hence, x = 2 and x = - 4 3 are the roots of the given equation.

Solve equation for x by factorisation 4 x - 3 = 5 2 x + 3 , x ≠ 0 , - 3 2

Given equation: 4 x - 3 = 5 2 x + 3 Now, ⇒ 4 - 3 x x = 5 2 x + 3 ⇒ 8 x - 6 x 2 + 12 - 9 x = 5 x ⇒ - 6 x 2 + 8 x - 9 x - 5 x + 12 = 0 ⇒ - 6 x 2 - 6 x + 12 = 0 ⇒ - 6 x 2 + x - 2 = 0 ⇒ x 2 + x - 2 = 0 By the method of splitting middle term we can write, ⇒ x 2 + 2 x - x - 2 = 0 ⇒ x ( x + 2 ) - 1 ( x + 2 ) = 0 ⇒ ( x + 2 ) ( x - 1 ) = 0 ⇒ x = - 2 or x = 1 Hence, x = - 2 and x = 1 are the roots of the given equation.

Solve equation for x by factorisation 3 x + 1 - 1 2 = 2 3 x - 1 , x ≠ - 1 , 1 3

The given equation is: 3 x + 1 - 1 2 = 2 3 x - 1 ⇒ 3 x + 1 - 2 3 x - 1 = 1 2 ⇒ 3 3 x - 1 - 2 x + 1 x + 1 3 x - 1 = 1 2 ⇒ 9 x - 3 - 2 x - 2 3 x 2 - x + 3 x - 1 = 1 2 ⇒ 7 x - 5 3 x 2 + 2 x - 1 = 1 2 ⇒ 3 x 2 + 2 x - 1 = 14 x - 10 ⇒ 3 x 2 - 12 x + 9 = 0 ⇒ x 2 - 4 x + 3 = 0 ⇒ x 2 - 3 x - x + 3 = 0 ⇒ x x - 3 - 1 x - 3 = 0 ⇒ x - 3 x - 1 = 0 ⇒ x - 3 = 0 or x - 1 = 0 ⇒ x = 3 or x = 1 Hence, x = 3 and x = 1 are the roots of the given equation.

Solve equation for x by factorisation 1 x - 3 + 2 x - 2 = 8 x , x ≠ 0 , 2 , 3

The given equation is: 1 x - 3 + 2 x - 2 = 8 x Taking LCM in the denominator of LHS, we get ⇒ x - 2 + 2 x - 3 x - 3 x - 2 = 8 x ⇒ x - 2 + 2 x - 6 x 2 - 2 x - 3 x + 6 = 8 x ⇒ 3 x - 8 x 2 - 5 x + 6 = 8 x Transposing x 2 - 5 x + 6 from LHS to RHS and x from RHS to LHS, we get ⇒ 8 x 2 - 40 x + 48 = 3 x 2 - 8 x ⇒ 5 x 2 - 32 x + 48 = 0 Factorising above quadratic equation by splitting the middle term, we get ⇒ 5 x 2 - 20 x - 12 x + 48 = 0 ⇒ 5 x x - 4 - 12 x - 4 = 0 ⇒ 5 x - 12 x - 4 = 0 ⇒ 5 x - 12 = 0 or x - 4 = 0 ⇒ 5 x = 12 or x = 4 ⇒ x = 12 5 or x = 4 Hence, x = 4 and x = 12 5 are the roots of the given equation.

Solve equation for x by factorisation x - 2 x - 3 + x - 4 x - 5 = 3 1 3 , x ≠ 3 , 5

The given equation is: x - 2 x - 3 + x - 4 x - 5 = 10 3 Taking LCM in the denominator, we get ⇒ x - 2 x - 5 + x - 3 x - 4 x - 3 x - 5 = 10 3 ⇒ x 2 - 5 x - 2 x + 10 + x 2 - 4 x - 3 x + 12 x 2 - 5 x - 3 x + 15 = 10 3 ⇒ 2 x 2 - 14 x + 22 x 2 - 8 x + 15 = 10 3 Taking 2 as a common factor in the numerator, we get ⇒ x 2 - 7 x + 11 x 2 - 8 x + 15 = 5 3 ⇒ 3 x 2 - 21 x + 33 = 5 x 2 - 40 x + 75 ⇒ 2 x 2 - 19 x + 42 = 0 ⇒ 2 x 2 - 12 x - 7 x + 42 = 0 ⇒ 2 x x - 6 - 7 x - 6 = 0 ⇒ x - 6 2 x - 7 = 0 ⇒ x - 6 = 0 or 2 x - 7 = 0 ⇒ x = 6 or x = 7 2 Hence, x = 6 and x = 7 2 are the roots of the given equation.

Solve equation for x by factorisation 3 x + 1 + 4 x - 1 = 29 4 x - 1 , x ≠ - 1 , 1 , 1 4

The given equation is: 3 x + 1 + 4 x - 1 = 29 4 x - 1 ⇒ 3 x - 1 + 4 ( x + 1 ) x - 1 ( x + 1 ) = 29 4 x - 1 ⇒ 3 x - 3 + 4 x + 4 x 2 - 1 = 29 4 x - 1 ⇒ 7 x + 1 x 2 - 1 = 29 4 x - 1 ⇒ 7 x + 1 4 x - 1 = 29 x 2 - 29 ⇒ 28 x 2 - 7 x + 4 x - 1 = 29 x 2 - 29 ⇒ x 2 + 3 x - 28 = 0 Now split the middle term ⇒ x 2 + 7 x - 4 x - 28 = 0 ⇒ x x + 7 - 4 x + 7 = 0 ⇒ x - 4 x + 7 = 0 ⇒ x - 4 = 0 or x + 7 = 0 ⇒ x = 4 or x = - 7 Hence, x = 4 and x = - 7 are the roots of the given equation.