# Determining interval solution of irrational inequality

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• Feb 18th 2006, 03:57 AM
DenMac21
Determining interval solution of irrational inequality
I have to solve irrational inequality:
$\sqrt x + \sqrt {x + 7} + 2\sqrt {x^2 + 7x} < 35 - 2x$

Inequality is defined for $0 \le x < 17.5$

Transforming inequality we get that:
$144x^2 - 2137x + 7569 > 0$
Solving this quadratic inequality we get that
$x_1 = 9 \wedge x_2 = \frac{{841}}{{144}} \approx 5.84$

So, solutions of beggining irrational inequality must be either
$0 \le x < 5.84$ or $9 < x < 17.5$.

In book stands that solution is $0 \le x < 5.84$, but
how do I know that this interval is solution?

I mean, how do I determine exactly which interval is solution?
• Feb 18th 2006, 05:09 AM
topsquark
Quote:

Originally Posted by DenMac21
I have to solve irrational inequality:
$\sqrt x + \sqrt {x + 7} + 2\sqrt {x^2 + 7x} < 35 - 2x$

Inequality is defined for $0 \le x < 17.5$

Transforming inequality we get that:
$144x^2 - 2137x + 7569 > 0$
Solving this quadratic inequality we get that
$x_1 = 9 \wedge x_2 = \frac{{841}}{{144}} \approx 5.84$

So, solutions of beggining irrational inequality must be either
$0 \le x < 5.84$ or $9 < x < 17.5$.

In book stands that solution is $0 \le x < 5.84$, but
how do I know that this interval is solution?

I mean, how do I determine exactly which interval is solution?

Pick a representative from each interval: For simplicity, I would choose x=5 and x=10 to test.

-Dan
• Feb 18th 2006, 10:05 AM
DenMac21
Quote:

Originally Posted by topsquark
Pick a representative from each interval: For simplicity, I would choose x=5 and x=10 to test.

-Dan

Is there another way to prove that?
I mean exact proof.
• Feb 18th 2006, 01:40 PM
topsquark
Quote:

Originally Posted by DenMac21
Is there another way to prove that?
I mean exact proof.

I'm not sure what you are asking. You solved the problem to the point where you have two disjoint solution sets. The way to find out which is true (it could even be that BOTH are true) is to pick a point in each set and see if it it solves your original problem. Assuming that the work that lead to the solution sets is correct, it isn't so much a matter of proving anything as much as it is verifying a solution. In this case, you have two intervals that are solutions sets, so by picking one number from each you can determine which interval solves your original inequality.

The point is whenever you have a situation where you have to manipulate an equation to solve it, you may be introducing extra solutions. When you come down to the solution set you always have to ask the question "Do all of my solutions satisfy the original equation?" The only way I know of to tell is to plug the solutions into the original equation and take a look. In this case you have a system where you've squared both sides of the equation a couple of times and that usually introduces extra solutions, so you need to check them at the end.

Is there some sort of theorem that will tell you which solution to pick? Sometimes. For example, I teach Physics, and it is a common side effect of solving the quadratic equation that one of the solutions is unphysical. You can usually tell just by looking at the solution set which solutions are the ones you want. Math, though, tends to be a bit more subtle. When in doubt, the simplest way to find out which solution is meaningful is to simply check your answer to the original problem.

-Dan
• Feb 18th 2006, 03:14 PM
ThePerfectHacker
Maybe you can express
$\sqrt{x}+\sqrt{x+7}+\sqrt{x^2+7x}<35-2x$
As,
$\sqrt{x}+\sqrt{x+7}+\sqrt{x}\sqrt{x+7}<35-2x$
Now, let $y=\sqrt{x}$ thus,
$y+\sqrt{y^2+7}+y\sqrt{y^2+y}<35-2y^2$
Thus,
$\sqrt{y^2+7}<\frac{35-2y^2-y}{y+1}$
Now you can square both sides and elimante the root.
Do not know if that helps because that leads to the quartic equation.
• Feb 18th 2006, 03:29 PM
ThePerfectHacker
Quote:

Originally Posted by DenMac21
Is there another way to prove that?
I mean exact proof.

There is nothing wrong with testing an interval.
Just pick one point and see if it satisfies the conditions.
It is exact and it is considered a proof.
• Feb 18th 2006, 03:59 PM
DenMac21
Ok, thanks!