# What do I do wrong when solving square inequality?

• October 23rd 2006, 01:50 PM
Val21
What do I do wrong when solving square inequality?
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
• October 23rd 2006, 02:48 PM
OReilly
Quote:

Originally Posted by Val21
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

When you did 2nd power of
$\sqrt {8 - x} \ge x - 2$ you forgot to do one other thing that is needed for all solutions of this irrational inequation.

These types of irrational inequations are solved like this:
$\begin{array}{l}
\sqrt {a(x)} \ge b(x) \\
\left( {a(x) \ge b^2 (x) \wedge b(x) \ge 0} \right) \vee \left( {a(x) \ge 0 \wedge b(x) \le 0} \right) \\
\end{array}
$

You have done one condition for solution:
$\sqrt {8 - x} \ge x - 2 \Leftrightarrow 8 - x \ge (x - 2)^2 \wedge x - 2 \ge 0$

But there is also second:

$\begin{array}{l}
\sqrt {8 - x} \ge x - 2 \Leftrightarrow 8 - x \ge 0 \wedge x - 2 \le 0 \\
x \le 8 \wedge x \le 2 \\
\end{array}
$

So, all $x \le 2$ are also solutions.

Of course, you have condition from original inequation that $x \ge - 5$ so solutions are indeed $- 5 \le x \le 4$
• October 23rd 2006, 05:48 PM
Soroban
Hello, Val21!

Is a puzzlement . . .

Quote:

I have this enequality: . $\sqrt{x+5} - \sqrt{8-x} \leq 1$

I solved this inequality: . $-1 \leq x \leq 4$

But the correct answer is: . $-5 \leq x \leq 4$

What do I do wrong?

I sketched a graph of the situation. .The domain is $[-5,\,8].$

We have the curve: . $y_1 \:=\:\sqrt{x + 5}$
This is the upper half of a right-opening parabola with vertex (-5, 0).

We have the curve: . $y_2 \:=\:\sqrt{8 - x}$
This is the upper half of a left-opening parabola with vertex (8, 0).

We want the x-values where the difference $y_1 - y_2 \:\leq \:1$
Code:

              |       *..    |              *       ::::::* |      .*       ::::::::|::::*::::       ::::::::*        *       ::::*  |          *       ::*    |            *       ::      |     --*-------+--------+-----*--     -5      |        4    8

And the solution is indeed: . $-5 \leq x \leq 4$