Radical Equations 2 - resolved

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Xearf_987

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Feb 14, 2006
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Problem 2:

2(sq.root(x)) - sq.root(4x - 3) = 1/ sq.root(4x-3)

Can someone please show me how to solve this one? At least show me how to do the first few steps. I have an idea of how to go about solving it but I'm not entirely sure. The attempt I made is sort of messy and confusing so I'd rather not post it.
 
Re: Radical Equations 2

Xearf_987 said:
Problem 2:

2(sq.root(x)) - sq.root(4x - 3) = 1/ sq.root(4x-3)

Can someone please show me how to solve this one? At least show me how to do the first few steps. I have an idea of how to go about solving it but I'm not entirely sure. The attempt I made is sort of messy and confusing so I'd rather not post it.
Click to expand...

Here's the first step I would take. Multiply both sides of the equation by sqrt(4x - 3):

sqrt(4x -3)* 2 sqrt(x) - sqrt(4x -3)*sqt(4x - 3) = sqrt(4x - 3)*1/sqrt(4x - 3)

sqrt(4x - 3)*2 sqrt(x) - (4x - 3) = 1

After that, it gets "messy and confusing for" me, too (not really.....) but I hope you can take it from here, and if not, that you can show us what you got.
 
Re: Radical Equations 2

Your equation:
2sqrt(x) - sqrt(4x - 3) = 1 / sqrt(4x - 3)
crisscross multiplication:
sqrt(4x - 3)[2sqrt(x) - sqrt(4x - 3)] = 1

2sqrt(4x^2 - 3x) - (4x - 3) = 1
2sqrt(4x^2 - 3x) - 4x + 3 = 1

2sqrt(4x^2 - 3x) = 4x - 2
sqrt(4x^2 - 3x) = 2x - 1

Now square both sides, then finish off.
 
Um, for my answer... I got - 4/ 11


Did I screw up? And should I show my work?

It doesn't seem to fit into the original equation, so would the answer be a null set?

S={ }
 
Ahhh.. I'll go ahead and post my work

[Original Problem]

2(sq.root(x)) - sq.root(4x - 3) = 1/ sq.root (4x - 3)

[Multiplied Each member by sq.root(4x - 3), like you said]

2(sq.root(4x^2 - 3x)) - 4x-3 = 1

[Added Like terms]

2(sq.root(4x^2 - 3x)) = 4x + 4

[Divided each side by 2]

sq.root(4x^2 - 3x) = 2x + 2

[Squared both sides]

4x^2 - 3x = 4x^2 + 8x + 4

[Added like terms]

-4 = 11x

x= -4/11

Did I screw up??
 
Minor point. A sign error. It is
2(sqrt(4x^2 - 3x)) - (4x - 3) = 1
which is
2(sqrt(4x^2 - 3x)) - 4x + 3 = 1
You should always try your solution in the original equation. -4/11 - 3 is negative so you cannot take the sqrt. That tells you something is wrong.
 
Well, on this assingment, the answer doesn't necessarily have to fit into the original equation. Some answers are null sets. But yes, I see where I made my mistake. Thank you very much!
 
Well... If your answer is that there is no answer, true, you can't put that into the original equation but if you say the answer is x=5 that does have to make the original equation true and you should ALWAYS try it.
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Gene
 
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