1. ## Solutions of quadratic equation

Hi having a bit of a mind blank trying to find the solutions to this quadratic equation, any hints would be invaluable

x^2 -(5/3)x +4/3=0

2. Originally Posted by offahengaway and chips
Hi having a bit of a mind blank trying to find the solutions to this quadratic equation, any hints would be invaluable

x^2 -(5/3)x +4/3=0
Have you tried using the formula ?
Note that there are no real solutions to this equation.

3. Yes, I keep getting math error on my calculator in the root part of the formula.

4. Originally Posted by offahengaway and chips
Yes, I keep getting math error on my calculator in the root part of the formula.
Run it complex mode.. The solution is complex.

5. Are there any fixed points on the graph of this equation, because that is what i am trying to answer

6. Originally Posted by offahengaway and chips
Are there any fixed points on the graph of this equation, because that is what i am trying to answer
The question is not clear.. What does fixed point of an equation mean?Fixed points are defined for functions (as far as I know).
Also usually when one says "solve an equation", one should specify the domain... Is x real?

So assuming we want x real, is the question asking for the fixed point of the function f(x) = x^2 -(5/3)x +4/3 ?

7. Yes, sorry, it is the fixed points of f(x)

8. Originally Posted by offahengaway and chips
Yes, sorry, it is the fixed points of f(x)
Then that means you have to solve f(x) = x.

This means you have to actually solve x^2 -(8/3)x +4/3=0. This equation can be safely graphed to get real solutions

9. x^2 -(5/3)x +4/3=0 is the equation from the function, unfortunately I have to find the answers algebraicly, that is where i am struggling.

10. Originally Posted by offahengaway and chips
x^2 -(5/3)x +4/3=0 is the equation from the function, unfortunately I have to find the answers algebraicly, that is where i am struggling.
First of all, do you understand how I got x^2 -(8/3)x +4/3=0 and secondly, do you know the quadratic formula?
I am not understanding where you are getting stuck..

11. I do know the quadratic formula, not too sure where you got the 8/3 from

12. Originally Posted by offahengaway and chips
I do know the quadratic formula, not too sure where you got the 8/3 from
You are asked to find fixed points of f(x) = x^2 -(5/3)x +4/3. Read this, to know the definition.

Now we know that a solution to f(x) = x is called a fixed point of f(x).
So find values of x for which f(x) = x.

f(x) = x implies x^2 -(5/3)x +4/3 = x, which implies x^2 -(8/3)x +4/3=0

Now hit it with the quadratic formula and get x. Do you get it now?

13. Sorry, I just cant see how the 8 comes into it

14. Originally Posted by offahengaway and chips
Sorry, I just cant see how the 8 comes into it

Just basic algebraic manipulations:

$x^2 - \frac53 x + \frac43 = x \implies \left(x^2 - \frac53 x + \frac43 \right) - x = 0 \implies x^2 - \left(\frac53 + 1 \right) x + \frac43$ $= 0 \implies x^2 - \frac83 x + \frac43 = 0
$

The only thing I have not explained is $1 + \frac53 = \frac83$.

15. AHAA, the penny has dropped thank you very much!!!!!!

so simple, such a daft mistake

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