JacksonJRB one year ago Solve for x in terms of y. y=x^2-4x+3

1. SolomonZelman

ok, I will show a similar problem.

2. JacksonJRB

ok

3. SolomonZelman

$$\large\color{black}{ \displaystyle y=2x^2-3x+5 }$$ $$\large\color{black}{ \displaystyle 0=2x^2-3x+5-y }$$ so a=2 b=-3 c=(5-y) $$\large\color{black}{ \displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(5-y)}}{2(2)} }$$ $$\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-8(5-y)}}{4} }$$ $$\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-40+8y}}{4} }$$ $$\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{-31+8y}}{4} }$$

4. JacksonJRB

So, i just use quadratic formula?

5. SolomonZelman

Yes

6. JacksonJRB

Okay, I'll try it. Thanks

7. SolomonZelman

ok

8. JacksonJRB

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

9. SolomonZelman

let me check ...

10. JacksonJRB

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

11. JacksonJRB

Simplifies to that I believe

12. SolomonZelman

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!

13. SolomonZelman

Good job, keep that up:)

14. JacksonJRB

Thanks for the help

15. SolomonZelman

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

16. JacksonJRB

There's another problem that involves fractions though

17. SolomonZelman

ok, ...

18. JacksonJRB

Little confused

19. JacksonJRB

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

20. SolomonZelman

you want x in terms of y?

21. JacksonJRB

yes

22. SolomonZelman

$$\large\color{black}{ \displaystyle y=\frac{x^2+1}{x^2-1} }$$ $$\large\color{black}{ \displaystyle y=\frac{x^2-1+2}{x^2-1} }$$ $$\large\color{black}{ \displaystyle y=\frac{x^2-1}{x^2-1}+\frac{2}{x^2-1} }$$ $$\large\color{black}{ \displaystyle y=1+\frac{2}{x^2-1} }$$ can you get it from there?

23. JacksonJRB

24. SolomonZelman

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

25. SolomonZelman

-1+2, is same as 1

26. JacksonJRB

Oh, I see now

27. JacksonJRB

I think I can finish it

28. SolomonZelman

29. JacksonJRB

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

30. SolomonZelman

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

31. JacksonJRB

You're right

32. JacksonJRB

Thanks

33. SolomonZelman

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.

34. SolomonZelman

yw in any case.