"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.
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.
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...
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
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 ?
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.