Question

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

Answer 1

Supposition:-

• let the no.s be x and y
• x²-y² = 45
• y² = 4x

According to the question:-

:$\implies$ $\sf x²-4x = 45$

:$\implies$ $\sf x²-4x-45 = 0$

:$\implies$ $\sf x²-(-5+9)x-45 = 0$

:$\implies$ $\sf x²+5x-9x-45=0$

:$\implies$ $\sf x(x+5)-9(x-5)=0$

:$\implies$ $\sf (x-9)(X+5) = 0$

:$\implies$ $\sf X = 9, -5$

but it is given that the no. is natural hence, -5 will be rejected and X = 9 will be accepted

so, X = 9 and y = 4x = 4×9 = 36

Answer 2

Required Answer:-

Given Statements:

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

To Find:

• The numbers.

Solution:

Let the numbers be x and y where x is the larger number, i.e.,

→ x > y

So,

→ "The difference of squares of the natural numbers is - 45"

This means,

→ x² - y² = 45 — (i)

Also,

→ "The square of the smaller number is four times the larger number."

This means,

→ y² = 4x (y is the smaller number and x is the larger)

Substituting the value of y² in (i), we get,

→ x² - 4x = 45

→ x² - 4x - 45 = 0

→ Now, we will solve the quadratic equation.

→ -9 × 5 = -45 and -9 + 5 = -4

So,

→ x² - 9x + 5x - 45 = 0

→ x(x - 9) + 5(x - 9) = 0

→ (x + 5)(x - 9) = 0

By zero product rule,

→ Either x + 5 = 0 or x - 9 = 0

→ x = 9, - 5

But x is a natural number. So, x cannot be negative.

→ x ≠ -5

→ x = 9

Now,

→ y² = 4x

→ y² = 4 × 9

→ y² = 36

→ y = √36

→ y = ±6

But y is also a natural number. It can't be negative.

→ y ≠ -6

→ y = 6

Hence,

→ x = 9

→ y = 6

★ Therefore, the numbers are 9 and 6.

Answer:

• The numbers are 9 and 6.