Radical Equation: sqrt(2x - sqrt(8x - 4)) = 1 + sqrt(x)

[miror]

i having a problem solving this equation

thanks!

 

[Deleted member 4993]

i having a problem solving this equation

thanks!

Hint: square both sides of the equation - simplify then square again ... tell us what you find....

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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[samir]

Hint: square both sides of the equation - simplify then square again ... tell us what you find....

What are your thoughts?

\(\displaystyle 2x-\sqrt{8x-4}=(1+\sqrt{x})^2\)

Now what?

 

[Deleted member 4993]

\(\displaystyle 2x-\sqrt{8x-4}=(1+\sqrt{x})^2\)

Now what?

\(\displaystyle 2x-\sqrt{8x-4}=(1+\sqrt{x})^2\)

\(\displaystyle 2x-\sqrt{8x-4} \ \ = \ \ 1 \ \ + \ \ 2*\sqrt{x} \ \ + \ \ x\)

\(\displaystyle x\ \ - \ \ 1 \ \ = \ \ 2*\sqrt{x} \ \ + \ \sqrt{8x-4}\)

square again and simplify .... square again and simplify .... till and untill you get rid of all the \(\displaystyle \sqrt{}\)

 

[samir]

\(\displaystyle 2x-\sqrt{8x-4}=(1+\sqrt{x})^2\)

\(\displaystyle 2x-\sqrt{8x-4} \ \ = \ \ 1 \ \ + \ \ 2*\sqrt{x} \ \ + \ \ x\)

\(\displaystyle x\ \ - \ \ 1 \ \ = \ \ 2*\sqrt{x} \ \ + \ \sqrt{8x-4}\)

square again and simplify .... square again and simplify .... till and untill you get rid of all the \(\displaystyle \sqrt{}\)

I don't get it. It's not getting any simpler.

\(\displaystyle \sqrt{2x-\sqrt{8x-4}}=1+\sqrt{x}\)

\(\displaystyle (\sqrt{2x-\sqrt{8x-4}})^2=(1+\sqrt{x})^2\)

\(\displaystyle 2x-\sqrt{8x-4}=(1+\sqrt{x})^2\)

\(\displaystyle 2x-\sqrt{8x-4}=1+2\sqrt{x}+x\)

\(\displaystyle (2x-\sqrt{8x-4})^2=(1+2\sqrt{x}+x)^2\)

\(\displaystyle 4x-4x\sqrt{8x-4}+8x-4=1+2\sqrt{x}+x+4x+2x\sqrt{x}+x+2x\sqrt{x}+x^2\)

Can it get any simpler than that? All I can see is more and more square roots, but of smaller quantities. Would you please post the complete solution? Or at least give me step by step instructions? What laws, rules or identities are you using?

 

[stapel]

I don't get it. It's not getting any simpler.

\(\displaystyle \sqrt{\strut 2x\, -\,\sqrt{\strut 8x\,-\,4\,}\,}\,=\,1\,+\,\sqrt{\strut x\,}\)

\(\displaystyle \left(\sqrt{\strut 2x\,-\,\sqrt{\strut 8x\,-\,4\,}\,}\right)^2\,=\,\left(1\,+\,\sqrt{\strut x\,}\right)^2\)

\(\displaystyle 2x\,-\,\sqrt{\strut 8x\,-\,4}\,=\,\left(1\,+\,\sqrt{\strut x\,}\right)^2\)

\(\displaystyle 2x\,-\,\sqrt{\strut 8x\,-\,4\,}\,=\,1\,+\,2\sqrt{\strut x\,}\,+\,x\)

You were given the above. Then, in reply, you proposed the following as the best next step:

\(\displaystyle \left(2x\,-\,\sqrt{\strut 8x\,-\,4\,}\right)^2\,=\,\left(1\,+\,2\sqrt{\strut x\,}\,+\,x\right)^2\)

No. Before you square again, simplify as much as you can, like you were shown in an earlier reply:

\(\displaystyle 2x\,-\,\sqrt{\strut 8x\,-\,4\,}\,=\,\left(1\,+\,\sqrt{\strut x\,}\right)^2\)

\(\displaystyle 2x\,-\,\sqrt{\strut 8x\,-\,4} \ \ = \ \ 1 \ \ + \ \ 2*\sqrt{\strut x\,} \ \ + \ \ x\)

\(\displaystyle x\ \ - \ \ 1 \ \ = \ \ 2*\sqrt{\strut x\,} \ \ + \ \sqrt{\strut 8x\,-\,4\,}\)

square again and simplify .... square again and simplify .... till and untill you get rid of all the \(\displaystyle \sqrt{}\)

You've now been given more than half of the solution, plus the steps for completing the solution. Please start with what you've been given, and then follow the method specified. Yes, this is messy; yes, this is long. Keep plugging away!