I have this enequality:

sqr(x+5)-sqr(8-x)<=1

1) I find the area of determination (sorry if I misuse math slang),

because the expression under roots cannot be negative

x+5>=0 x>=-5 and

8-x>=0 x<=8

which means -5<=x<=8

2) Now I work with my inequation

sqr(x+5)-sqr(8-x)<=1

sqr(x+5)<=1+sqr(8-x)

Now because both left and right sides of the inequality ARE positive

(because squre roots are positive) I can make them both in 2nd power and the truthfulness of the statement won't change.

(sqr(x+5))^2<=(1+sqr(8-x))^2

x+5<=1+2*sqr(8-x)+8-x

x+5-1-8+x<=2*sqr(8-x)

2x-4<=2*sqr(8-x)

x-2<=sqr(8-x)

(x-2)^2<=(sqr(8-x))^2

x^2-4x+4<=8-x

x^2-3x-4<=0

I solve this enequality and find area of x-s which lie into the determination area (however this is called

-1<=x<=4

which I thought was the answer two the exersice,but

the correct answer (thank's God we have answers in the math book)

is -5<=x<=4

What do I do wrong?

Thanks