Question

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

Hint:

When a problem specifies "positive numbers," always reject the negative root obtained from the quadratic equation.

Answer 1

Understanding the Problem:
Let the two positive numbers be $x$ and $y$.
It is given that the sum of their squares is 100 and one number exceeds the other by 2.

Forming Equations:
1) $x^2 + y^2 = 100$
2) $x = y + 2$

Substitution Method:
Substitute $x = y + 2$ into the first equation:
$(y + 2)^2 + y^2 = 100$

Expand:
$y^2 + 4y + 4 + y^2 = 100$
$2y^2 + 4y + 4 = 100$

Simplify:
$2y^2 + 4y - 96 = 0$
Divide by 2:
$y^2 + 2y - 48 = 0$

Factorization:
$(y + 8)(y - 6) = 0$

So, $y = -8$ or $y = 6$.
Since the numbers are positive, $y = 6$.

Now,
$x = y + 2 = 6 + 2 = 8$.

Final Answer:
The two positive numbers are 6 and 8.