The sum of squares of two numbers is 68 and t | vedclass

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Question

(A) Let the two numbers be $a$ and $b$.Given that the sum of their squares is $a^2 + b^2 = 68$.Also,the square of their difference is $(a - b)^2 = 36$

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let the two numbers be $a$ and $b$.Given that the sum of their squares is $a^2 + b^2 = 68$.Also,the square of their difference is $(a - b)^2 = 36$.We know the algebraic identity: $(a - b)^2 = a^2 + b^2 - 2ab$.Substituting the given values into the identity:$36 = 68 - 2ab$.Rearranging the terms to solve for $2ab$:$2ab = 68 - 36$.$2ab = 32$.Dividing by $2$:$ab = 16$.Therefore,the product of the two numbers is $16$.

The sum of squares of two numbers is $68$ and the square of their difference is $36$. The product of the two numbers is

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