square root ques.

**shannu82** · September 17th 2010, 02:33 PM

Find all the real values of x such that
1/((sqrt(2x-3))^2) > 1/4

I just want to know how to do this ques.
So far i have done the following:
4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)
x<19/2 (isolated x...)

Any help would be greatly appreciated.

**e^(i*pi)** · September 17th 2010, 02:49 PM

Originally Posted by shannu82

Find all the real values of x such that
1/((sqrt(2x-3))^2) > 1/4

I just want to know how to do this ques.
So far i have done the following:
4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)
x<19/2 (isolated x...)

Any help would be greatly appreciated.

Good start, but where did you get 19/2 from? I got 7/2 from adding 3 to both sides to eliminate the -3

You have: $4 > 2x-3$

Add 3 to both sides: $7 > 2x$

Can you finish?

From what I understand you are squaring before taking the square root so the domain is all the real numbers except 3/2

**shannu82** · September 17th 2010, 03:01 PM

Originally Posted by e^(i*pi)

Good start, but where did you get 19/2 from? I got 7/2 from adding 3 to both sides to eliminate the -3

You have: $4 > 2x-3$

Add 3 to both sides: $7 > 2x$

Can you finish?

From what I understand you are squaring before taking the square root so the domain is all the real numbers except 3/2

Thanx for replying, ya... i for some reason had squared the 4.... I got 7/2. So I just wanna confirm that the answer is x <7/2 is the final ans? I figure that even if you apply the 3/2 sqrt and sqr cancel each other out!
~Priya

**e^(i*pi)** · September 17th 2010, 03:06 PM

Yes, that's correct

**Krizalid** · September 17th 2010, 03:55 PM

Originally Posted by shannu82

Find all the real values of x such that
1/((sqrt(2x-3))^2) > 1/4

I just want to know how to do this ques.
So far i have done the following:
4 > 2x-3 ( multiplied left side by 4 and right side by (sqrt(2x-3))^2). ( sqrt x square only 2x-3 remains)

$\sqrt{x^2}$ and $\big(\sqrt x\big)^2$ are different, the first one equals $|x|$ which holds for any $x$ and the second one only holds for $x\ge0,$ so on your question, the inequality does actually require that $2x-3>0,$ and then by algebra it follows that $x<\dfrac72,$ but since $x>\dfrac32,$ the solution for the original inequality is $\dfrac32<x<\dfrac72.$

**shannu82** · September 17th 2010, 04:18 PM

I just want to make sure i understand correctly, if the above question was changed from (√(2x-3))^2 to √((2x-3)^2) would it be x<7/2

**Krizalid** · September 18th 2010, 01:23 PM

no, it's not correct.

now in this case we're dealing with $\dfrac1{|2x-3|}>\dfrac14,$ which can be written as $|2x-3|<4,$ we can do this since $|2x-3|>0,$ well it can be zero but makes no sense in this case, so $-4<2x-3<4\implies-\dfrac12<x<\dfrac72.$

but this is not the solution set since $x\ne\dfrac32,$ so the solution set is actually $\displaystyle\left] -\frac{1}{2},\frac{3}{2} \right[\cup \left] \frac{3}{2},\frac{7}{2} \right[.$

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