Question

who can give me two distinct methods solution set for inequality y^2>x^2 please

Answer 1

Factor, factor, factor...

\(y^{2} = x^{2}\)

\(y^{2} - x^{2} = 0\)

\((y-x)(y+x) = 0\)

y = x or y = -x These are the broken borders. We have only to pick the proper sides of these borders.

Answer 2

thanks so much but what is the set solution for inequality y^2>x^2 algebraically please

Answer 3

y^2 > x^2
y^2 - x^2 > 0
(y-x)(y+x) > 0
y > x or y < -x

Answer 4

thank you so much but i need the set solution for inequality y^2>x^2 algebraically please

Answer 5

thank you so i found (y-x)(y+x)>0 that main y-x>0 and y+x>0 or y-x<0and y-x<0 so is that wrong ?

Answer 6

(y-x)(y+x) > 0
The product of two factors are positive IF BOTH factors are positive
or if both factors are negative.
(y-x)(y+x) > 0 IF y-x > 0 AND y+x > 0
Make the assumption x is positive.
Then y-x > 0 implies y > x. If x is positive then y > x will make y+x > 0.
Therefore, the product will be positive.
So y > x will make the product positive.

Second possibility: y - x < 0 AND y + x < 0
Use similar reasoning to figure what y will make the product positive.

Then, consider x is negative and follow similar reasoning to figure out what y will make the product positive.