# solving an equation which contains x and sqrtx

• Jan 5th 2009, 03:36 AM
Sebastian de Vries
solving an equation which contains x and sqrtx
Techniques of Differentiation
When $3x+4\sqrt{x}-1=0$ then why is $\sqrt{x}=\frac{-2+\sqrt{7}}{3}?$
It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $x$ wouldn't do any good since I would get $x\sqrt{x}$. 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
• Jan 5th 2009, 03:40 AM
Constatine11
Quote:

Originally Posted by Sebastian de Vries
Techniques of Differentiation
When $3x+4\sqrt{x}-1=0$ then why is $\sqrt{x}=\frac{-2+\sqrt{7}}{3}?$
It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with $x$ wouldn't do any good since I would get $x\sqrt{x}$. 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 $y=\sqrt{x}$ to get the quadratic:

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

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

.
• Jan 5th 2009, 03:42 AM
craig
Hmm, when I do this I get:

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

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

My above answer was done using the quadratic forumula by the way
• Jan 5th 2009, 03:47 AM
mr fantastic
Quote:

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

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

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

$\sqrt{28} = 2 \sqrt{7}$ ....
• Jan 5th 2009, 03:49 AM
craig
(Speechless) how did I miss that one lol.
• Jan 5th 2009, 05:04 AM
Sebastian de Vries
Thanks.