Question

25. The sum of squares of two numbers is 145. If square root of one number is 3
then find the other number
​

Answer 1

Let the two numbers be x , y and Sum of squares of two numbers is 145 so as :-

$\\ {\pmb{\sf{ x^2+y^2 = 145 \ \ \ \ \ \ \ \ \cdots(1) }}}$

Now, we've Square root of ${\sf{ 1^{st} }}$ number that is ${\sf{ 3 }}$: We can Square both sides to make it Easier to use as:-

$\colon\mapsto{\sf{ ( \sqrt{x} ) = 3 }} \\ \\ \colon\mapsto{\sf{ ( \sqrt{x} )^2 = 3^2 }} \\ \\ \colon\mapsto{\sf{ x = 9 }}$

So, ${\sf{ 1^{st} }}$ number is 9

From Equation (1), we can find the value of ${\sf{ 2^{nd} }}$ number as:-

$\colon\mapsto{\pmb{\sf{ x^2+y^2 = 145 }}} \\ \\ \colon\mapsto{\sf{ (9)^2 + y^2 = 145 }} \\ \\ \colon\mapsto{\sf{ 81 + y^2 = 145 }} \\ \\ \colon\mapsto{\sf{ y^2 = 145 - 81 }} \\ \\ \colon\mapsto{\sf{ y^2 = 64 }} \\ \\ \colon\mapsto{\sf{ y = \sqrt{64} \mapsto 8 }}$

Hence, ${\sf{ 2^{nd} }}$ number will be 8.