Solve the equation 3 x 2 + 10 x + 7 3 = 0 by factorisation.

We are given to solve the equation 3 x 2 + 10 x + 7 3 = 0 by factorisation. We shall use the method of splitting the middle term for the factorisation. Thus, we have, 3 x 2 + 10 x + 7 3 = 0 ⇒ 3 x 2 + 3 x + 7 x + 7 3 = 0 ⇒ 3 x x + 3 + 7 x + 3 = 0 ⇒ x + 3 3 x + 7 = 0 ⇒ x + 3 = 0 or 3 x + 7 = 0 ⇒ x = - 3 or x = - 7 3 . Therefore, the solutions of the given equation are - 3 and - 7 3 .

Solve the equation 6 x - 2 x - 1 = 1 x - 2 by factorisation.

We are given to solve the equation 6 x - 2 x - 1 = 1 x - 2 by factorisation. First, we shall simplify the given equation, convert it into a quadratic equation in its standard form and then, use the method of splitting the middle term for the factorisation. Thus, we have, 6 x - 2 x - 1 = 1 x - 2 ⇒ 6 x - 1 - 2 x x x - 1 = 1 x - 2 ⇒ 6 x - 6 - 2 x x 2 - x = 1 x - 2 ⇒ 4 x - 6 x 2 - x = 1 x - 2 ⇒ 4 x - 6 x - 2 = x 2 - x ⇒ 4 x 2 - 8 x - 6 x + 12 - x 2 + x = 0 ⇒ 3 x 2 - 13 x + 12 = 0 ⇒ 3 x 2 - 9 x - 4 x + 12 = 0 ⇒ 3 x x - 3 - 4 x - 3 = 0 ⇒ x - 3 3 x - 4 = 0 ⇒ x - 3 = 0 or 3 x - 4 = 0 ⇒ x = 3 or x = 4 3 . Therefore, the solutions of the given equation are 3 and 4 3 .

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 .