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• May 14th 2009, 04:26 AM
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
• May 14th 2009, 04:35 AM
Isomorphism
Quote:

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.
• May 14th 2009, 04:37 AM
offahengaway and chips
Yes, I keep getting math error on my calculator in the root part of the formula.
• May 14th 2009, 04:37 AM
Isomorphism
Quote:

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.
• May 14th 2009, 04:38 AM
offahengaway 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 AM
Isomorphism
Quote:

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 ?
• May 14th 2009, 05:03 AM
offahengaway and chips
Yes, sorry, it is the fixed points of f(x)
• May 14th 2009, 05:15 AM
Isomorphism
Quote:

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 ;)
• May 14th 2009, 05:29 AM
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.
• May 14th 2009, 05:32 AM
Isomorphism
Quote:

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..
• May 14th 2009, 05:39 AM
offahengaway and chips
I do know the quadratic formula, not too sure where you got the 8/3 from
• May 14th 2009, 05:43 AM
Isomorphism
Quote:

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?
• May 14th 2009, 05:51 AM
offahengaway and chips
Sorry, I just cant see how the 8 comes into it
• May 14th 2009, 05:57 AM
Isomorphism
Quote:

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

(Surprised)

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 $1 + \frac53 = \frac83$.
• May 14th 2009, 06:07 AM
offahengaway and chips
AHAA, the penny has dropped thank you very much!!!!!!

so simple, such a daft mistake
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