# Thread: Checking my work on this inequality

1. ## Checking my work on this inequality

My answer disagrees with the solution given in the book. I want to make sure I'm right.

x^2 <= y^2 +XY - 1 If X=3, find the value(s) of y that satisfy the equation.

The given answer is -5 <= y <= 2, but that's not right, because for y=1,

9 <= 1 + 3 - 1 is clearly not true.

My solution is:

9 <= y^2 +3y -1

0 <= y^2 +3y -10

0<= (y + 5) (y - 2)

Y+5=0 y = -5

y-2=0 y =2

Are the roots and also solutions.

For y < -5, the product of (y+5) (y-2) (try -6) is +, therefore it's part of the solution.

For -5<y<2, the product of (y+5)(y-2) (try 1) is -, therefore it's not part of the solution.

For y>2, the product of (y+5)(y-2) (try 3) is positive, therefore it's part of the solution.

y <= -5
y >= 2

2. Originally Posted by fcabanski
x^2 <= y^2 +XY - 1 If X=3, find the value(s) of y that satisfy the equation.
The given answer is -5 <= y <= 2, but that's not right, because for y=1,
My solution is:
9 <= y^2 +3y -1
0 <= y^2 +3y -10
0<= (y + 5) (y - 2)
y <= -5
y >= 2
This solution is correct to the problem as you have written it.
The book’s answer is correct if the problem were $x^2 {\color{red}\ge} y^2+xy-1$.

3. Thanks for the verification.