Question

One number is 6 more than anouther number. the sum of their squares is 90

Answer 1

Let x = greater number
y = smaller number

(1)
`x = y + 6`
(1)
`y = x-6`

(2)
`x^(2) + y^(2) = 90`
We'll substitute y in (1) to (2)
(2)
`x^(2) + (x-6)^(2) = 90`

`x^(2) + (x^(2) - 12x + 36) = 90`

`x^2 + x^2 -12x + 36 - 90 = 0`

`2x^2 - 12x -54 = 0`

`2(x^2 - 6x - 27) = 0`

`2(x - 9)(x + 3) = 0`
x - 9 = 0 or x + 3 = 0
x = 9 x = -3
and
y = x - 6 y = x - 6
y = 9 - 6 y = -3 - 6
y = 3 y = -9

Therefore, the two numbers can be 9 and 3 or -3 and -9.