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# Inequality

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Senior Manager
Joined: 12 Mar 2010
Posts: 357

Kudos [?]: 270 [0], given: 87

Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34

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01 Feb 2012, 04:23
Solve the inequality: sqrt(2x+3) > x
link to the source (example 5 in http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398420&section=2.2)

I have done it as follows:

Leg1:

=> sqrt(2x+3) > x
=> 2x+3 > x^2
=> x^2 -2x -3 < 0
=> -1< x < 3

Leg 2:

2x+3 has to be positive
=> 2x+3 > 0
=> x > -3/2

Considering both the solution is -1< x < 3

Where did I go wrong?

Kudos [?]: 270 [0], given: 87

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Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129047 [0], given: 12187

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01 Feb 2012, 04:53
bsaikrishna wrote:
Solve the inequality: sqrt(2x+3) > x
link to the source (example 5 in http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398420&section=2.2)

I have done it as follows:

Leg1:

=> sqrt(2x+3) > x
=> 2x+3 > x^2
=> x^2 -2x -3 < 0
=> -1< x < 3

Leg 2:

2x+3 has to be positive
=> 2x+3 > 0
=> x > -3/2

Considering both the solution is -1< x < 3

Where did I go wrong?

If we approach the way you propose:

Expression under the square root can not be negative: $$2x+3\geq{0}$$ --> $$x\geq{-\frac{3}{2}}$$. Notice that for $$-\frac{3}{2}\leq{x}<0$$ LHS>0>RHS, so for this range the given inequality holds true.

Next, when you square both parts (squaring an inequality is a tricky thing) you'l get: $$x^2-2x-3<0$$ --> $$-1<x<3$$. Combining both ranges: $$-\frac{3}{2}\leq{0}<3$$.

I'd personally find the range $$-\frac{3}{2}\leq{x}<0$$ first and then consider the cases for $$x\geq{0}$$. We can safely square both part of the inequality if we know that both parts of an inequality are non-negative (so our case) --> $$x^2-2x-3<0$$ --> $$-1<x<3$$, as we consider the range $$x\geq{0}$$ then: $$0\leq{x}<{3}$$. Combine two ranges: $$-\frac{3}{2}\leq{0}<3$$.

GENERAL RULE:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.
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Kudos [?]: 129047 [0], given: 12187

Senior Manager
Joined: 12 Mar 2010
Posts: 357

Kudos [?]: 270 [0], given: 87

Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34

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01 Feb 2012, 05:52
Yes, I got where I went wrong. I shouldn't square on both the sides here because 'x' can take negative values too.

So,let us say I am not going to square it both the sides.

Then for a problem like this sqrt( 2x + 3) > x

all I know is 2x+3 >=0, implies x belongs to [-3/2, infinity)

So, how do I solve it from there on?

Kudos [?]: 270 [0], given: 87

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Posts: 41892

Kudos [?]: 129047 [0], given: 12187

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01 Feb 2012, 05:57
bsaikrishna wrote:
Yes, I got where I went wrong. I shouldn't square on both the sides here because 'x' can take negative values too.

So,let us say I am not going to square it both the sides.

Then for a problem like this sqrt( 2x + 3) > x

all I know is 2x+3 >=0, implies x belongs to [-3/2, infinity)

So, how do I solve it from there on?

As I explained thins in my previous post, first you find the range $$-\frac{3}{2}\leq{x}<0$$ and then consider the case for $$x\geq{0}$$. At this point you CAN square as both part of the inequality are positive and you'll get: $$x^2-2x-3<0$$ --> $$-1<x<3$$, as we consider the range $$x\geq{0}$$ then: $$0\leq{x}<{3}$$. Combine two ranges: $$-\frac{3}{2}\leq{0}<3$$.
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02 Feb 2012, 01:24
bsaikrishna wrote:
Solve the inequality: sqrt(2x+3) > x
link to the source (example 5 in http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398420&section=2.2)

I have done it as follows:

Leg1:

=> sqrt(2x+3) > x
=> 2x+3 > x^2
=> x^2 -2x -3 < 0
=> -1< x < 3

Leg 2:

2x+3 has to be positive
=> 2x+3 > 0
=> x > -3/2

Considering both the solution is -1< x < 3

Where did I go wrong?

The problem is the way you approached this question. You have obviously done some such questions to have an idea of what to do. But a little bit of clarity of thought is required here.

When you see: $$\sqrt{(2x+3)} > x$$, you say to yourself, "Ok, this inequality will behave differently under different circumstances. There are 2 cases: x is negative or x is positive (or 0) (this is because you can square the inequality if both sides are non negative)

Case 1: x is negative i.e. x < 0
$$\sqrt{(2x+3)}$$ will be always positive so this inequality will hold for all negative x. But the constraint here is that 2x+3 should remain positive because sqrt of a negative number is not defined.
So, 2x + 3 > 0
x > -3/2
Mind you x must be greater than -3/2 if it is negative so this inequality holds for -3/2 < x < 0

Case 2: x is positive or 0 i.e. x >= 0
In that case, x can take only those values for which this inequality: $$\sqrt{(2x+3)} > x$$ holds.
Here, both sides of the inequality are non negative so you can square it.
You get (2x + 3) > x^2
which gives you -1 < x < 3
But mind you, you get this solution only when you assume that x is positive.
So this inequality holds for 0 <= x < 3

Now you combine these two solutions (in bold) and you get -3/2 < x < 3
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Re: Inequality   [#permalink] 02 Feb 2012, 01:24
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# Inequality

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