Question

Please help me with this problem?
Find the values of x:
tx^2+ux=v

Answer 1

Do you know about the quadratic formula?

Answer 2

From the formula \[\Large ax^2+bx+c\]
the value of x can be calculated as such
\[\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 3

I hate abstract math, but this prob is not very hard

Answer 4

This problem has the same sort of setup except that the a, b and c look different

Answer 5

yes, to find the c, first, subtract v from both sides.

Answer 6

bu you might rather want to complete the square

Answer 7

That also works, do you know how to complete the square @musicmakeslife

Answer 8

don't want to type all of this abstract nonsense though

Answer 9

Complete the square

\(tx^2+ux=v\\tx^2+ux-v=0\\t(x^2+\frac{u}{t})^2-v=0\\t(x+\frac{u}{2t})^2=v+2(\frac{u}{2t})^2\)

If you keep going like this its going to look allot like \(x=\frac{-u\pm\sqrt{u^2-4t(-v)}}{2t}\)

Answer 10

SO in other words @zzr0ck3r ,it would have been easier to do from the beginning?

Answer 11

And to complete the square you first need to divide everything by t since it can't have any coefficient other than 1 on it

Answer 12

I don't know what would be easier... When I first saw it, I thought complete the square, but in doing that I realize I am just deriving the quadratic formula that you posted.

I never divided the t away, I just factored it out of the first two terms.

Answer 13

you could, but then you would end up writing more fractions, and I hate doing that in latex...

Answer 14

Ok I kind of understand, I knew it had something to do with the Quadratic formula @doulikepiecauseidont and also completing the square @zzr0ck3r, I apologize for taking so long to respond, my network is acting up. I'm confused about completing the square though.

Answer 15

In essence, I suggested using the quadratic formula because there are no numbers to go with it. completing the square will get you a lot of unnecessary steps that you may or may not want, as @zzr0ck3r shows and says he didn''t like fractions. Quad formula gives you the answer in one step, mostly

Answer 16

I did set it up in quadratic formula in the first place, but I'm unsure how to simplify from there on. This problem does come with a list of possible answers, would you like me to post them? @doulikepiecauseidont

Answer 17

@doulikepiecauseidont

Answer 18

@eric_d could you help me please?

Answer 19

I thought @doulikepiecauseidont and a few others had explained to you..! ??
@musicmakeslife

Answer 20

They explained to use the quadratic equation and how to work out the completeing the square, but I'm still confused, I'm sorry @eric_d

Answer 21

\[tx^2+ux-v=0\]
\[tx^2+uv+(\frac{ u }{ 2 })^2-(\frac{ u }{ 2 })^2-v=0\]

Answer 22

\[(tx+\frac{ u }{ 2 })^2-\frac{ u^2 }{ 4 }-v=0\]

Answer 23

\[(tx+\frac{ u }{ 2 })^2 - \frac{ u^2 }{ 4 }- \frac{ 4v }{ 4 } = 0\]
\[(tx^2+\frac{ u }{ 2 })^2-\frac{ u^2-4v }{ 4 } = 0\]

Answer 24

@musicmakeslife

Answer 25

You mean this...?
@musicmakeslife

Answer 26

I got it @eric_d thanks for the help and sorry for the late response. :)

Answer 27

You're welcome @musicmakeslife !
~It's alrite...