Find the values that x2-y2 can take.

Question

-3<x≤4

 

-4<y<-1

 

Find the values that x²-y² can take. 

Answer 1

First factor your given equation: x^2 - y^2 = (x+y)*(x-y)

 

We have the x values of  -2, -1, 0, 1, 2, 3, and 4. Notice I did not include -3 on the list as x is simply greater than -3, NOT greater than or equal to -3.

 

We have the y values of -3, and -2 (assuming we're just using integers). Notice that y cannot equal -4 or -1 as you don't have a less than or equal to sign; you simply have a less than sign.

 

So now we can just plug in our numbers... start with x=-2 and then continue

 

(-2+(-3))*(-2-(-3)) = (-2-3)*(-2+3) = (-5)*(1) = -5

(-2+(-2))*(-2-(-2)) = (-2-2)*(-2+2) = (-4)*(0) = 0

 

(-1+(-3))*(-1-(-3)) = (-1-3)*(-1+3) = (-4)*(2) = -8

(-1+(-2))*(-1-(-2)) = (-1-2)*(-1+2) = (-3)*(1) = -3

 

(0+(-3))*(0-(-3)) = (0-3)*(0+3) = (-3)*(3) = -9

(0+(-2))*(0-(-2)) = (0-2)*(0+2) = (-2)*(2) = -4

 

(1+(-3))*(1-(-3)) = (1-3)*(1+3) = (-2)*(4) = -8

(1+(-2))*(1-(-2)) = (1-2)*(1+2) = (-1)*(3) = -3

 

(2+(-3))*(2-(-3)) = (2-3)*(2+3) = (-1)*(5) = -5

(2+(-2))*(2-(-2) = (2-2)*(2+2) = (0)*(4) = 0

 

(3+(-3))*(3-(-3)) = (3-3)*(3+3) = (0)*(6) = 0

(3+(-2))*(3-(-2)) = (3-2)*(3+2) = (-1)*(5) = -5

 

(4+(-3))*(4-(-3)) = (4-3)*(4+3) = (1)*(7) = 7

(4+(-2))*(4-(-2)) = (4-2)*(4+2) = (2)*(6) = 12

 

So we have the values, -9, -8, -5, -4, -3, 0, 7, and 12 for our outputs. I hope this was helpful!