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

Step 1: Solving OR (B):
1. Let the numbers be $x$ and $x+2$.
2. Equation: $x^2 + (x+2)^2 = 100$.
3. $x^2 + x^2 + 4x + 4 = 100 \implies 2x^2 + 4x - 96 = 0$.
4. Divide by 2: $x^2 + 2x - 48 = 0$.
5. Factorize: $(x+8)(x-6) = 0$.
6. Since numbers are positive, $x = 6$. The numbers are 6 and $6+2=8$.
Step 2: Final Answer (OR):
The two positive numbers are 6 and 8.