SOLUTION: one positive number is 4 more than another. The sum of their squares is 40. what are the numbers ?

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Question 246343: one positive number is 4 more than another. The sum of their squares is 40. what are the numbers ?
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
let n and m =the numbers
n=4+m
n^2+m^2=40
substitute
(m+4)^2+m^2=40
m^2+8m+16+ m^2=40
2m^2+8m+16-40=0
2m^2+8m-24=0
divide by 2
m^2+4m-12=0
we can only use the positive answer
n=2
m=2+4=6
so we have 2+4=6 for the other number
2^2+6^2=40
4+36=40
40=40
check
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=64 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 2, -6. Here's your graph:
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