"If f(x)=kx^2+8x+k-6 determine the values of k that will give f(x) two zeroes."

Please show me how to do this, as well as answer it. It is very important and any help is great. Thanks.

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- Jun 16th 2008, 11:12 AMFORKFinding Values of K (Zeroes)
"If f(x)=kx^2+8x+k-6 determine the values of k that will give f(x) two zeroes."

Please show me how to do this, as well as answer it. It is very important and any help is great. Thanks. - Jun 16th 2008, 11:19 AMMoo
Hello,

Think "discriminant", $\displaystyle \Delta=b^2-4ac$ in $\displaystyle ax^2+bx+c$.

If $\displaystyle \Delta<0$, there is no zero within the real numbers.

If $\displaystyle \Delta=0$, there is one zero.

If $\displaystyle \Delta>0$, there are 2 zeroes.

(Wink) - Jun 16th 2008, 11:21 AMFORK
- Jun 16th 2008, 11:24 AMMoo
- Jun 16th 2008, 11:26 AMFORK
- Jun 16th 2008, 11:36 AMMoo
Huum... I don't really like it... but I just hope you will

*learn*it...

$\displaystyle f(x)=kx^2+8x+k-6=\underbrace{k}_{a}x^2+\underbrace{8}_{b}x+\under brace{(k-6)}_{c}$

$\displaystyle \Delta=b^2-4ac=(8)^2-4*k*(k-6)=64-4k(k-6)$

You can see that there is only one situation where you can have 2 zeroes : $\displaystyle \Delta>0$

$\displaystyle 64-4k(k-6)>0$

$\displaystyle 4k^2-24k-64<0$

$\displaystyle k^2-6k-16<0$

$\displaystyle (k+2)(k-8)<0$

Try to solve... - Jun 16th 2008, 11:42 AMFORK
- Jun 16th 2008, 11:50 AMFORK
- Jun 16th 2008, 11:56 AMMoo
No, there are the values where (x+2)(x-8)<0, so that will be sets of values.

I'll give you a hint : a product of two factors is negative if and only if the two are of different sign.

This means that you have to solve two things :

- x+2>0 AND x-8<0

- x+2<0 AND x-8>0

One of the situation will not be possible (Wink) - Jun 16th 2008, 12:02 PMFORK
- Jun 16th 2008, 12:13 PMMoo
You can do it in two ways :)

The one I tried to describe with solving k+2>0 etc...

And another one, which is simpler, that's why I changed the method :

$\displaystyle k^2-6k-16<0$

The roots are -2 and 8.

I don't know if you have learnt that if a polynomial $\displaystyle ax^2+bx+c$ has roots $\displaystyle x_1$ and $\displaystyle x_2$ (and $\displaystyle x_1<x_2$), then we can have the sign of the polynomial :

- sign of a if $\displaystyle x \in ]-\infty ~,~x_1] \cup [x_2 ~,~ +\infty[$

- opposit sign of a if $\displaystyle x \in [x_1 ~,~ x_2]$

Here, a=1, what can you conclude ? :) - Jun 16th 2008, 12:27 PMFORK
- Jun 16th 2008, 12:36 PMjanvdl
You have:

$\displaystyle (k + 2)(k - 8) < 0$

So we have the points $\displaystyle -2$ and $\displaystyle 8$.

$\displaystyle k^2$ is positive so the part to the right of $\displaystyle 8$ is positive, the part to the left of $\displaystyle -2$ will also be positive.

We want the part smaller than $\displaystyle 0$.

Those numbers lie between $\displaystyle -2$ and $\displaystyle 8$, but the two aforementioned numbers are not included. - Jun 16th 2008, 12:45 PMFORK
- Jun 16th 2008, 12:46 PMMoo