Question

Hey guys, I have another question, this one involves the quadratic formula. Trying to find all solutions for x^2-5=squareroot of 3x using the quadratic formula

Answer 1

\[x ^{2}-5=\sqrt{3x}\]

Answer 2

Was this an assigned problem? The quadratic formula isn't really applicable to this case.

Answer 3

Solve the equation 2−5= root 3x using the quadratic formula. Approximate solutions to three decimals places.

Answer 4

I know I would put it into A+B+C form like
\[x ^{2}-\sqrt{3x}-5\]
But when I try to plus it into the equation it turns into a mess

Answer 5

You sure it's not supposed to be 2x-5 = sqrt(3x)

Answer 6

sqrt(3x) = sqrt(3)*sqrt(x)

Answer 7

or maybe
\[\sqrt{3}x\]?

Answer 8

I don't see how that helps him, Estudiar.

Answer 9

I think you're miscopying the problem somehow. Either the x isn't supposed to be under the radical or that first term isn't x^2 but just x.

Answer 10

because if you square both sides you are going to get a 4th degree polynomial with a linear term (degree 1) and no quadratic formula will help you there

Answer 11

Yes. You could use the quartic formula! Hurrah!

Answer 12

Ohhh yes! YOur right Smooth, it is \[\sqrt{3}\]x

Answer 13

Thanks :p I make silly mistakes, I should be able to work it out now.

Answer 14

Ah, well then that is simple. It becomes x^2 - sqrt(3)*x -5 = 0.

Answer 15

Thanks for the help. I will go work it out.

Answer 16

whew. now use
\[x^2-\sqrt{3}x-5=0\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=1,b=-\sqrt{3},c=-5\]