Question

I already solved this problem. I just need help with the HINT part.

The sum of two numbers is 25 and the sum of their squares is 313. Find the number. (HINT: LET ONE OF THE NUMBERS BE X. EXPRESS THE OTHER NUMBER IN TERMS OF X.)

Answer 1

`x+y=25 \\ y=25-x~~Other~number~in~terms~of~x \\ \\ x^2+y^2=313 \\ x^2+2xy+y^2=313+2xy \\ (x+y)^2=313+2x(25-x) \\ (25^2-313)/(2) =25x-x^2 \\ x^2-25x+156 \\ \\ \Delta =(-25)^2-4* 1* 156 \\ \Delta =625-624 \\ √(\Delta)=1 \\ \\ x= (25 \pm 1)/(2) \\ x= \left \{ {{x_1=13} \atop {x_2=12}} \right.`

The number "x" is given by:

`y= \left \{ {{y_1=25-13 \rightarrow~y_1=12} \atop {y_1=25-12 \rightarrow ~y_1=13}} \right.`

Therefore, we have:


`\boxed {S=({12,13})} \rightarrow~The~curly~braces~does~not~want~appear~here.`