Q2 of 227 Page 107

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.

Let ‘x’ be the larger number and ‘y’ be the smaller number.


x2 – y2 = 45 …(1)


y2 = 4x …(2)


Now, put the value of y2 in equation (1).


x2 – 4x = 45


x2 – 4x 45 = 0


x2– 9x + 5x – 45 = 0


x(x – 9) + 5(x 9) = 0


(x + 5)(x 9) =0


x + 5 = 0 or x – 9 = 0


x = –5 or x = 9


Here positive 9 only admissible. From this we need to find the value of y for that we are going to aplly this value in the second equation.


y2 = 4 x


y2 = 4 × 9


y2 = 36


y = 36


y = ± 6


Here, positive 6 only admissible.


Therefore, the required numbers are 6 and 9.


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