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

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

Mathematics
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ok, I will show a similar problem.
ok
\(\large\color{black}{ \displaystyle y=2x^2-3x+5 }\) \(\large\color{black}{ \displaystyle 0=2x^2-3x+5-y }\) so a=2 b=-3 c=(5-y) \(\large\color{black}{ \displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(5-y)}}{2(2)} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-8(5-y)}}{4} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-40+8y}}{4} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{-31+8y}}{4} }\)

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Other answers:

So, i just use quadratic formula?
Yes
Okay, I'll try it. Thanks
ok
\[x=\frac{ 4+/- \sqrt{4+4y} }{ 2 }\]
let me check ...
\[x=2+/-\sqrt{1+y}\]
Simplifies to that I believe
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!
Good job, keep that up:)
Thanks for the help
you have done it all yourself, i just pointed:)
There's another problem that involves fractions though
ok, ...
Little confused
\[y=\frac{ x^2+1 }{ x^2-1 }\]
you want x in terms of y?
yes
\(\large\color{black}{ \displaystyle y=\frac{x^2+1}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=\frac{x^2-1+2}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=\frac{x^2-1}{x^2-1}+\frac{2}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=1+\frac{2}{x^2-1} }\) can you get it from there?
Why did you add 2?
I didn't, I rewrote 1 as 2-1
-1+2, is same as 1
Oh, I see now
I think I can finish it
ok, go ahead, but take your time...
\[x=\sqrt{\frac{ 2 }{ y-1 }+1}\]
yes (I think though you need the \(\pm\) there)
You're right
Thanks
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.
yw in any case.

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