"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, 12:12 PMFORKFinding 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, 12:19 PMMoo
Hello,

Think "discriminant", in .

If , there is no zero within the real numbers.

If , there is one zero.

If , there are 2 zeroes.

(Wink) - Jun 16th 2008, 12:21 PMFORK
- Jun 16th 2008, 12:24 PMMoo
- Jun 16th 2008, 12:26 PMFORK
- Jun 16th 2008, 12:36 PMMoo
Huum... I don't really like it... but I just hope you will

*learn*it...

You can see that there is only one situation where you can have 2 zeroes :

Try to solve... - Jun 16th 2008, 12:42 PMFORK
- Jun 16th 2008, 12:50 PMFORK
- Jun 16th 2008, 12:56 PMMoo
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, 01:02 PMFORK
- Jun 16th 2008, 01:13 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 :

The roots are -2 and 8.

I don't know if you have learnt that if a polynomial has roots and (and ), then we can have the sign of the polynomial :

- sign of a if

- opposit sign of a if

Here, a=1, what can you conclude ? :) - Jun 16th 2008, 01:27 PMFORK
- Jun 16th 2008, 01:36 PMjanvdl
You have:

So we have the points and .

is positive so the part to the right of is positive, the part to the left of will also be positive.

We want the part smaller than .

Those numbers lie between and , but the two aforementioned numbers are not included. - Jun 16th 2008, 01:45 PMFORK
- Jun 16th 2008, 01:46 PMMoo