Question

The sum of squares of two positive numbers is 100. If one number exceeds the other by 2, find the numbers.

Answer 1

One number excess the other by 2 

Let those numbers be x and x + 2

According to question,

x² + (x + 2)² = 100

⇒ x² + x² + 2² + 2(x)(2) = 100

⇒ x² + x² + 4 + 4x = 100

⇒ 2x² + 4x – 96 = 0

⇒ x² + 2x – 48 = 0   ...[On dividing both sides by 2] 

On comparing with standard form,

ax² + bx + c = 0

a = 1, b = 2, c = –48

Roots are given by `(-b ± sqrt(b^2 - 4ac))/(2a)`

So, Roots = `(-2 ± sqrt(2^2 - 4 xx 1 xx (-48)))/(2 xx 1)`

= `(-2 ± sqrt(4 + 192))/2`

= `(-2 ± 14)/2`

6, –8   ...(∵ –8 is rejected)

∴ Numbers are 6 and (6 + 2) or 6 and 8.