# Finding Values of K (Zeroes)

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• Jun 16th 2008, 11:12 AM
FORK
Finding 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 AM
Moo
Hello,

Quote:

Originally Posted by FORK
"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.

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

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

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

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

(Wink)
• Jun 16th 2008, 11:21 AM
FORK
Quote:

Originally Posted by Moo
Hello,

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

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

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

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

(Wink)

Thanks, i do know that part though. Can you show me step by step how to solve this type of problem? Please
• Jun 16th 2008, 11:24 AM
Moo
Quote:

Originally Posted by FORK
Thanks, i do know that part though. Can you show me step by step how to solve this type of problem? Please

Trust me it would be better for you to know how to solve it...

First of all, define a,b and c, like I wrote in $ax^2+bx+c$.
Then, find $\Delta$.

See if you can reach this part ^^
• Jun 16th 2008, 11:26 AM
FORK
Quote:

Originally Posted by Moo
Trust me it would be better for you to know how to solve it...

First of all, define a,b and c, like I wrote in $ax^2+bx+c$.
Then, find $\Delta$.

See if you can reach this part ^^

Im at the end of my semester and its the last time i will need it, so at this point it will not really matter. Plus, after i see youre example I will understand it, since I do not have any examples in my notes.
• Jun 16th 2008, 11:36 AM
Moo
Huum... I don't really like it... but I just hope you will learn it...

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

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

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

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

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

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

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

Try to solve...
• Jun 16th 2008, 11:42 AM
FORK
Quote:

Originally Posted by Moo
Huum... I don't really like it... but I just hope you will learn it...

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

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

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

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

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

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

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

Try to solve...

Thanks so much, I understand it now. Now i am just writing this so please do not close the thread.
• Jun 16th 2008, 11:50 AM
FORK
Quote:

Originally Posted by Moo
Huum... I don't really like it... but I just hope you will learn it...

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

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

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

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

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

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

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

Try to solve...

So then the values of k that will give two zeros are -2, and 8 right?
• Jun 16th 2008, 11:56 AM
Moo
Quote:

Originally Posted by FORK
So then the values of k that will give two zeros are -2, and 8 right?

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 PM
FORK
Quote:

Originally Posted by Moo
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)

Ok still not understanding.

Can you just please list the steps from the point you left at like before? That was very helpful and at least let me understand how to get to that point.
• Jun 16th 2008, 12:13 PM
Moo
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 :

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

The roots are -2 and 8.

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

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

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

Here, a=1, what can you conclude ? :)
• Jun 16th 2008, 12:27 PM
FORK
Quote:

Originally Posted by Moo
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 :

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

The roots are -2 and 8.

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

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

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

Here, a=1, what can you conclude ? :)

This keeps getting even more confusing, is all i need to know that the roots are -2 and 8, therefore the values of k that will give 2 zeros are -2 and 8?
• Jun 16th 2008, 12:36 PM
janvdl
You have:

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

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

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

We want the part smaller than $0$.

Those numbers lie between $-2$ and $8$, but the two aforementioned numbers are not included.
• Jun 16th 2008, 12:45 PM
FORK
Quote:

Originally Posted by janvdl
You have:

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

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

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

We want the part smaller than $0$.

Those numbers lie between $-2$ and $8$, but the two aforementioned numbers are not included.

Oh ok i get it, so then -2,-1,0,1,2,3,4,5,6,7,8 are all possible values of k that will give f(x) two zeroes?
• Jun 16th 2008, 12:46 PM
Moo
Quote:

Originally Posted by FORK
Oh ok i get it, so then -2,-1,0,1,2,3,4,5,6,7,8 are all possible values of k that will give f(x) two zeroes?

If k=-2 or k=-8, the $\Delta$ will be equal to 0, which we don't want (see post #2)

Plus, k can be a non-integer, so it will be all real numbers k : $-2
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