Question

A positive real number is 6 less than another. If the sum of the squares of the two numbers is 38, then find the numbers?

Answer 1

Let the positive number be `x` & other number be `x+6` `\quad \forall \ x>0`

As per given condition, we have sum of squares of both numbers equal to 38

`\therefore x^2+(x+6)^2=38`

`x^2+x^2+12x+36=38`

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

`x^2+6x-1=0`

Now, using quadratic formula to find roots of above quadratic formula as follows

`x=\frac{-6\pm\sqrt{6^2-4(1)(-1)}}{2(1)}`

`x=\frac{-6\pm2\sqrt{10}}{2}`

`x=-3\pm\sqrt{10}`

But, `x>0` hence `x=-3+\sqrt{10}`

`x=\sqrt{10}-3\approx 0.16227766016837952`