Question

If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

• A. -32
• B. 48
• C. 64
• D. 16

Answer 1

The Correct Option is B

If the sum of the squares of two numbers \(a\) and \(b\) is 97, we express this mathematically as:\(a^2 + b^2 = 97\).

The product of the numbers is \(ab\). We know from the identity:
\((a+b)^2 = a^2 + b^2 + 2ab\) 

Substituting \(a^2 + b^2 = 97\), we have:
\((a+b)^2 = 97 + 2ab\)

Rearranging gives:
\(2ab = (a+b)^2 - 97\)

This equation tells us that if \(p\) is the possible product \(ab\), then:
\(2p = (a+b)^2 - 97\)

Therefore:\((a+b)^2 = 2p + 97\)

\(a+b\) must be a real number, implying that \((a+b)^2\) should be a non-negative perfect square.

Let's test each option:

For \(p = -32\):
\(2(-32) + 97 = 33\) (not a perfect square)

For \(p = 48\):
\(2(48) + 97 = 193\) (not a perfect square)

For \(p = 64\):
\(2(64) + 97 = 225 = 15^2\) (perfect square)

For \(p = 16\):
\(2(16) + 97 = 129\) (not a perfect square)

Therefore, the option that cannot be the product is \(48\) because 193 is not a perfect square.