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

Printable View

- May 14th 2009, 04:26 AMoffahengaway and chipsSolutions 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 - May 14th 2009, 04:35 AMIsomorphism
Have you tried using the formula ?

Note that there are no real solutions to this equation. - May 14th 2009, 04:37 AMoffahengaway and chips
Yes, I keep getting math error on my calculator in the root part of the formula.

- May 14th 2009, 04:37 AMIsomorphism
- May 14th 2009, 04:38 AMoffahengaway and chips
Are there any fixed points on the graph of this equation, because that is what i am trying to answer

- May 14th 2009, 04:59 AMIsomorphism
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 ? - May 14th 2009, 05:03 AMoffahengaway and chips
Yes, sorry, it is the fixed points of f(x)

- May 14th 2009, 05:15 AMIsomorphism
- May 14th 2009, 05:29 AMoffahengaway 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.

- May 14th 2009, 05:32 AMIsomorphism
- May 14th 2009, 05:39 AMoffahengaway and chips
I do know the quadratic formula, not too sure where you got the 8/3 from

- May 14th 2009, 05:43 AMIsomorphism
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? - May 14th 2009, 05:51 AMoffahengaway and chips
Sorry, I just cant see how the 8 comes into it

- May 14th 2009, 05:57 AMIsomorphism
(Surprised)

Just basic algebraic manipulations:

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

$

The only thing I have not explained is $\displaystyle 1 + \frac53 = \frac83$. - May 14th 2009, 06:07 AMoffahengaway and chips
AHAA, the penny has dropped thank you very much!!!!!!

so simple, such a daft mistake