Question 487579: Please help me solve this equation.
Two integers have a sum of -4. The sum of their squares is 40. What are the two integers?
Thanks! Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! x+y=-4
x=-4-y…………(1)
x^2+y^2=40……….(2)
Put the value of x in (2)
(-4-y)^2+y^2=40
[(-4)+(-y)]^2+y^2=40
16+2(-4)(-y)+ y^2 +y^2=40
16+8y+ y2 +y^2=40
2y^2+8y+16=40
2y^2+8y=40-16
2y^2+8y=40-16
2y^2+8y=24
2(y^2+4y)=24
Divide by 2 both side
2(y^2+4y)/2=24/2
y^2+4y=12
y^2+4y-12=0
y^2+6y-2y-12=0
y(y+6)-2(y+6)=0
(y-2)(y+6)=0
y-2=0 or y+6=0
y=2 or y=-6
put the value of y=2 or y=-6 in (1)
x=-4-2 or x=-4-(-6)
x=-6 or x=-4+6
x=-6 or x=2
so two numbers are 2 and -6
