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Question

(A) Let the two numbers be $a$ and $b$.Given that the sum of their squares is $a^2 + b^2 = 80$.Also,the square of their difference is $(a - b)^2 = 36$

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(A) Let the two numbers be $a$ and $b$.Given that the sum of their squares is $a^2 + b^2 = 80$.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:$36 = 80 - 2ab$.Rearranging the terms to solve for $2ab$:$2ab = 80 - 36 = 44$.Dividing by $2$ to find the product $ab$:$ab = 22$.Therefore,the product of the two numbers is $22$.

The sum of squares of two numbers is $80$ and the square of their difference is $36$. The product of the two numbers is

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