Question

√x + √2x =1

Answer 1

Solve for x over the real numbers:
\(\sqrt x+\sqrt 2 \sqrt x = 1\)
Eliminate the square roots on both sides.
Square both sides:
\((\sqrt x + \sqrt 2 \sqrt x)^2 = 1\)
Expand and collect in terms of x.
\((\sqrt x+\sqrt 2 \sqrt x )^2 = (3+2 \sqrt2) x:\)
\((3+2 \sqrt 2) x = 1\)
Solve for x.
Divide both sides by \(3+2 \sqrt 2\):
Which leaves you with...?

Answer 2

@DangerousJesse
Your approach is correct but the answer is wrong:
http://www.wolframalpha.com/input/?i=solve+%E2%88%9Ax+%2B+%E2%88%9A2x+%3D1

Answer 3

\[\sqrt x+\sqrt 2 \sqrt x = 1\] can be factored as follows:\[(1+\sqrt{2})\sqrt{x}=1\]

Answer 4

@Adjax if you would have tried to work through it, rather than simply reading what Wolfram said, you would see that I got the same answer :)

Answer 5

...which can be solved for Sqrt(x) and then for x itself.

Answer 6

factoring, when it is equal to one, would only help if the factors are perfect square (which would be known as completing the square) .

Answer 7

\(\normalsize\color{blue}{ \sqrt{x}+\sqrt{2x}=1\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ x+2x+2x\sqrt{2}=1\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ 3x+2x\sqrt{2}=1\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ 3x-1=-2x\sqrt{2}\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ 9x^2-6x+1=8x\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ 9x^2-14x+1=0\LARGE\color{white}{ \rm │ }}\)

Answer 8

Now, solve for x using the quadratic formula.

Answer 9

ok lets carry throught the working:


dividing both sides:

therefore:
\[x = \frac{ 1 }{ 3 + 2 \sqrt{2} }\]
.......
which isn't write answer...I haven't just saw wolframs answer and posted that....I do got through yar working and thats what I'm trying to say

Answer 10

this is the answer:
\[x = 3-2 \sqrt{2} \approx 0.17157\]

@SolomonZelman : Nice approach ..your answer is very near to that..but isn't using quadratic approach here gives some unwanted solution too..I mean t isn't itss........I think you've got what I'm trying to say...

Answer 11

I would still suggest that one of the easier approaches to solving this equation for x is to factor the left side FIRST, obtaining (as before) \[(1+\sqrt{2})\sqrt{x}=1,\]...which, when solved for Sqrt(x), results in \[\sqrt{x}=\frac{ 1 }{ 1+\sqrt{2} }.\] Square both sides and solve for x. Finally, check EVERY result (so that any extraneous root may be identified and eliminated).

Answer 12

Note that the best form of the final solution has no radical or radicals in the denominator. We need to rationalize the denominator to get to that point.

Answer 13

Please, @Adjax, don't tell people that they are wrong because Wolfram said so, when you haven't even tried to work through it :) Try working through the problem, and you will see that all of these answers are accurate interpretations.

Answer 14

I got throught your working and I'm not saying that you are wrong ..your approach is correct..uh..got it..sorry your answer is right..I"m solving with something other.....what a huge mistake I made
+1 Jesse

Answer 15

Adjax, you probably mean unwanted is not exact... but the exact solution will not change from the method we use.

Answer 16

I mad a blunder mistake there in haste..therfore sorry @DangerousJesse ..this approach gives 2 or more solution to x...

Answer 17

*again mistake ther:solutions depends on the degrees there

Answer 18

@allisuarez00 : This problem was your posting. How about joining the discussion? Have you any comments or questions?