# square root ques.

• Sep 17th 2010, 03:33 PM
shannu82
square root ques.
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.
• Sep 17th 2010, 03:49 PM
e^(i*pi)
Quote:

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
• Sep 17th 2010, 04:01 PM
shannu82
Quote:

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
• Sep 17th 2010, 04:06 PM
e^(i*pi)
Yes, that's correct
• Sep 17th 2010, 04:55 PM
Krizalid
Quote:

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
• Sep 17th 2010, 05:18 PM
shannu82
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
• Sep 18th 2010, 02:23 PM
Krizalid
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

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