Question

Solve for x in terms of y.
y=x^2-4x+3

Answer 1

ok, I will show a similar problem.

Answer 2

ok

Answer 3

\(\large\color{black}{ \displaystyle y=2x^2-3x+5 }\)
\(\large\color{black}{ \displaystyle 0=2x^2-3x+5-y }\)

so
a=2
b=-3
c=(5-y)

\(\large\color{black}{ \displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(5-y)}}{2(2)} }\)

\(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-8(5-y)}}{4} }\)

\(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-40+8y}}{4} }\)

\(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{-31+8y}}{4} }\)

Answer 4

So, i just use quadratic formula?

Answer 5

Yes

Answer 6

Okay, I'll try it. Thanks

Answer 7

ok

Answer 8

\[x=\frac{ 4+/- \sqrt{4+4y} }{ 2 }\]

Answer 9

let me check ...

Answer 10

\[x=2+/-\sqrt{1+y}\]

Answer 11

Simplifies to that I believe

Answer 12

y=x^2-4x+3
0=x^2-4x+3-y

\(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{(-4)^2-4(1)(3-y)}}{2} }\)

\(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{16-4(3-y)}}{2} }\)

\(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{4(4-(3-y))}}{2} }\)

\(\large\color{black}{ \displaystyle x=\frac{4\pm2\sqrt{4-(3-y)}}{2} }\)

\(\large\color{black}{ \displaystyle x=2\pm\sqrt{4-(3-y)} }\)

\(\large\color{black}{ \displaystyle x=2\pm\sqrt{4-3+y} }\)

\(\large\color{black}{ \displaystyle x=2\pm\sqrt{1+y} }\)

Very Nice!

Answer 13

Good job, keep that up:)

Answer 14

Thanks for the help

Answer 15

you have done it all yourself, i just pointed:)

Answer 16

There's another problem that involves fractions though

Answer 17

ok, ...

Answer 18

Little confused

Answer 19

\[y=\frac{ x^2+1 }{ x^2-1 }\]

Answer 20

you want x in terms of y?

Answer 21

yes

Answer 22

\(\large\color{black}{ \displaystyle y=\frac{x^2+1}{x^2-1} }\)

\(\large\color{black}{ \displaystyle y=\frac{x^2-1+2}{x^2-1} }\)

\(\large\color{black}{ \displaystyle y=\frac{x^2-1}{x^2-1}+\frac{2}{x^2-1} }\)

\(\large\color{black}{ \displaystyle y=1+\frac{2}{x^2-1} }\)

can you get it from there?

Answer 23

Why did you add 2?

Answer 24

I didn't, I rewrote 1 as 2-1

Answer 25

-1+2, is same as 1

Answer 26

Oh, I see now

Answer 27

I think I can finish it

Answer 28

ok, go ahead, but take your time...

Answer 29

\[x=\sqrt{\frac{ 2 }{ y-1 }+1}\]

Answer 30

yes (I think though you need the \(\pm\) there)

Answer 31

You're right

Answer 32

Thanks

Answer 33

But, I mean without the \(\pm\) you have still achieved the task. You solved for x in terms of y. You might not need ± because (-)² or (+)² is all 1.

Answer 34

yw in any case.