# inequalities

• Jun 8th 2008, 02:48 AM
devanlevin
inequalities
seemingly simple problem i just cant seem to solve

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

x>=5
x<=8

because of the square roots, i can assume that all numbers are possitive so i can open up by multiplying by the power of 2

x+5-2*sqrt(x+5)(8-x)+8-x<=1
therefore
12<=2*sqrt(x+5)(8-x)
divide by 2 and then again power of 2-to get rid of the sqrt
36<=3x+40-x^2

now i solve it and i get -1<=X<=4 every time but the solution is -5<=x<=4

what am i doing wrong??
• .

the answer is any x between -5 and 4, it is the right answer but i dont know how to get to it...

i can only get from -1 to 4, but if you put any x till -5 it works, how do i solve this
• Jun 8th 2008, 10:21 AM
o_O
$36 \leq -x^{2} + 3x + 40$
$x^{2} - 3x - 4 \leq 0$
$(x - 4)(x + 1) \leq 0$

Your roots would be x = 4, x = -1 (These are the values that should theoretically satisfy the equality part of the inequality). However, if you plug in x = -1 to your original inequality, you would see that:
$\sqrt{5+(-1)} - \sqrt{8-(-1)} = \sqrt{4} - \sqrt{9} \neq 1$

So the only root you consider is x = 4. Choose a value higher than 4 and one below 4 to see which region would give you a number less than 1, keeping in mind that x is only defined over the interval $-5 \leq x \leq 8$
• Jun 8th 2008, 10:29 AM
Krizalid
Quote:

Originally Posted by o_O

However, if you plug in x = -1 to your original inequality, you would see that:
$\sqrt{1+5} - \sqrt{8-1} = \sqrt{6} - \sqrt{7} \neq 1$

:eek: Sure?

Having $-5\le x\le8$ and $-1\le x\le4,$ solution set is given by $[-5,8]\cap[-1,4]$ which gives the expected answer.
• Jun 8th 2008, 10:57 AM
o_O
Ah whoops, plugged in x = 1 instead of -1.