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

$64$

$48$

$-32$

$16$

✅ Correct Option: 1

Let the two numbers be $a$ and $b$ .

Given: $a^2 + b^2 = 97$

We need to find which value cannot be their product $ab$ .

To find the range of possible values for $ab$ , we use the identity:

$(a - b)^2 = a^2 + b^2 - 2ab$

Since $(a - b)^2 \geq 0$ for any real numbers, we have:

$a^2 + b^2 - 2ab \geq 0$

Substituting $a^2 + b^2 = 97$ :

$97 - 2ab \geq 0$

$97 \geq 2ab$

$ab \leq \dfrac{97}{2} = 48.5$

The maximum value $ab = 48.5$ is achieved when $(a - b)^2 = 0$

The product $ab$ must satisfy $ab \leq 48.5$ .

Therefore, any value greater than $48.5$ cannot be their product. Only such value from the options is $64$

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