Solve the equation by factorisation: 2 3 x 2 - 1 3 x = 1

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

Solve the equation by factorisation: 1 7 3 x - 5 2 = 28

The given equation is, 1 7 3 x - 5 2 = 28 We know that, ( a - b ) 2 = a 2 - 2 a b + b 2 Thus, 1 7 9 x 2 - 30 x + 25 = 28 Multiplying both sides by 7 , we get, ⇒ 9 x 2 - 30 x + 25 = 196 ⇒ 9 x 2 - 30 x + 25 - 196 ⇒ 9 x 2 - 30 x - 171 = 0 Dividing both sides by 3 , we get, ⇒ 3 x 2 - 10 x - 57 = 0 Factorising above quadratic equation by splitting the middle term, we get ⇒ ⇒ 3 x 2 - 19 x + 9 x - 57 = 0 ⇒ x 3 x - 19 + 3 3 x - 19 = 0 ⇒ 3 x - 19 x + 3 = 0 ⇒ 3 x - 19 = 0 or x + 3 = 0 ⇒ 3 x = 19 or x = - 3 ⇒ x = 19 3 or x = - 3 Hence, x = 19 3 and x = - 3 are the roots of the given equation.

Find two consecutive odd positive integers, sum of whose squares is 394 .

Let one of the number be x which is an odd positive integer then the other number be x + 2 . Then according to question, we have x 2 + x + 2 2 = 394 ⇒ x 2 + x 2 + 4 x + 4 = 394 ⇒ 2 x 2 + 4 x - 390 = 0 ⇒ x 2 + 2 x - 195 = 0 ⇒ x 2 + 15 x - 13 x - 195 = 0 ⇒ x x + 15 - 13 x + 15 = 0 ⇒ x - 13 x + 15 = 0 ⇒ x - 13 = 0 or x + 15 = 0 ⇒ x = 13 [ x ≠ - 15 , since we have let x to be is an odd positive integer] The other number is x + 2 = 13 + 2 = 15 Hence, the required numbers are 13 , and 15

The difference of the squares of two positive numbers is 45 . The square of the smaller number is four times the larger number. Find the numbers.

Consider, the greater number be x and the smaller number be y . According to the question: x 2 - y 2 = 45 … … … ( 1 ) y 2 = 4 x … … … ( 2 ) From ( 1 ) and ( 2 ) , we get: x 2 - 4 x = 45 ⇒ x 2 - 4 x - 45 = 0 ⇒ x 2 - 9 x + 5 x - 45 = 0 ⇒ x ( x - 9 ) + 5 ( x - 9 ) = 0 ⇒ ( x - 9 ) ( x + 5 ) = 0 ⇒ x - 9 = 0 or x + 5 = 0 ⇒ x = 9 ( ∵ x is a natural number) Putting the value of x in equation ( 2 ) , we get: y 2 = 36 ⇒ y = 6 Hence, the required numbers are x = 9 and y = 6 .