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Inequality

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Inequality [#permalink] New post 01 Feb 2012, 03:23
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

But the answer is: [-3/2,3)

Where did I go wrong?
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Re: Inequality [#permalink] New post 01 Feb 2012, 03:53
Expert's post
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

But the answer is: [-3/2,3)

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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Re: Inequality [#permalink] New post 01 Feb 2012, 04:52
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?
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Re: Inequality [#permalink] New post 01 Feb 2012, 04:57
Expert's post
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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Inequality [#permalink] New post 02 Feb 2012, 00:24
Expert's post
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

But the answer is: [-3/2,3)

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, 00:24
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