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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

There is a number consisting of two digits. The digit in the units place is twice that in the tens place,and if $2$ is subtracted from the sum of the digits,the difference is equal to $1/6^{\text{th}}$ of the number. The number is:

$A$ number $N$ is a positive three-digit number. If $x$ is in its hundred's place and $y$ is in its unit's place,then the number $N - 100x - y$ is always divisible by

The last digit,that is,the digit in the unit's place of the number $[(57)^{25}-1]$ is

The sum of three consecutive odd numbers is $20$ more than the first of these numbers. What is the middle number?

If the sum of two numbers is multiplied by each number separately,the products so obtained are $247$ and $114$. The sum of the numbers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module