# Thread: solving an equation which contains x and sqrtx

1. ## solving an equation which contains x and sqrtx

Techniques of Differentiation
When $\displaystyle 3x+4\sqrt{x}-1=0$ then why is $\displaystyle \sqrt{x}=\frac{-2+\sqrt{7}}{3}?$
It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $\displaystyle x$ wouldn't do any good since I would get $\displaystyle x\sqrt{x}$. Same if I would quadratize the equation by itself.
I think that they used the quadratic formula somehow but I don't see how.
Obviously the problem is the combination of

2. Originally Posted by Sebastian de Vries
Techniques of Differentiation
When $\displaystyle 3x+4\sqrt{x}-1=0$ then why is $\displaystyle \sqrt{x}=\frac{-2+\sqrt{7}}{3}?$
It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $\displaystyle x$ wouldn't do any good since I would get $\displaystyle x\sqrt{x}$. Same if I would quadratize the equation by itself.
I think that they used the quadratic formula somehow but I don't see how.
Obviously the problem is the combination of
Put $\displaystyle y=\sqrt{x}$ to get the quadratic:

$\displaystyle 3y^2+4y-1=0$

Now use the quadratic formula and you get two solutions, one of which is negative and so is a spurious solution since $\displaystyle y=\sqrt{x}\ge0$, the remaining positive root gives the solution.

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3. Hmm, when I do this I get:

$\displaystyle \sqrt{x}=\frac{-4\pm\sqrt{28}}{6}$

Not sure if this is something I've missed or what?

My above answer was done using the quadratic forumula by the way

4. Originally Posted by craig
Hmm, when I do this I get:

$\displaystyle \sqrt{x}=\frac{-4\pm\sqrt{28}}{6}$

Not sure if this is something I've missed or what?
$\displaystyle \sqrt{28} = 2 \sqrt{7}$ ....

5. how did I miss that one lol.

6. Thanks.