Question

The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Solve the problem specifically step by step !

Answer 1

Let the numbers be x and y where x>y so,

x^2 - y^2 = 120.....(1)

y^2 = 2 x.....(2)

Putting (2) in (1), we get

x^2 - 2x = 120

x^2 - 2x - 120 = 0

x^2 - 12x + 10x - 120 = 0

x(x -12) +10(x - 12) = 0

(x-12)(x+10) = 0

So x can be 12 or -10, but we are given that the square of smaller number is twice the larger number, therefore the square of any number cant be negative,

Thus x cant take value of -10

So x = 12, so y^2 = 2 (12)

y^2 = 24

So y = √24 or - √24

y = 2√6 or -2√6

Hence the numbers are 12 and 2√6 or 12 and -2√6

Answer 2

The answer is given below :

Let us consider that the numbers are p and q with p > q.

Then, by the given conditions,

Difference of the squares of p and q = 120

=> p² - q² = 120 .....(i)

and

Square of the smaller number = Twice the greater number

=> q² = 2p .....(ii)

Putting q² = 2p in (i), we get

p² - 2p = 120

=> p² - 2p + 1 = 120 + 1

=> (p - 1)² = 121

=> (p - 1)² = 11²

=> p - 1 = ± 11

=> p = 1 ± 11

Thus, we get

p = 1 + 11 and, p = 1 - 11

i.e., p = 12 and p = -10

When, p = 12, from (ii), we get

q² = 2 × 12

=> q² = 24

=> q² = 2² × 6

=> q = ± 2√6

Again, when p = -10, from (ii), we get

q² = 2 × (-10)

=> q² = -20

=> q² = 2² × 5 × i², where i = √(-1) and i = -1

=> q = ± 2√5 i

Hence, the required numbers are

(12, ± 2√6) and (-10, ± 2√5 i).

[Note : For lower classes, only calculate the value of q with p = 12 only, because complex number i = √(-1) may not be in syllabus.]

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