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§ion=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 < 0Case 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 < 3Now you combine these two solutions (in bold) and you get -3/2 < x < 3

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Karishma

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