1. Help me find k

a. the equation 3x^2 + 9x=17+6kx have roots that have equal magnitude but opposite in signs.
b. the graph of y=x^2+kx+k+8 intersects the x axis at two distinct points.

2. Originally Posted by hirano
a. the equation 3x^2 + 9x=17+6kx have roots that have equal magnitude but opposite in signs.
b. the graph of y=x^2+kx+k+8 intersects the x axis at two distinct points.

(a)

$\displaystyle 3x^2+9x=17+6kx\Rightarrow\ 3x^2+9x-6kx-17=0$

$\displaystyle \displaystyle\ ax^2+bx+c=0\Rightarrow\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\displaystyle 3x^2+(9-6k)x+(-17)=0$

For part (a), you require $\displaystyle \pm(value)$, so the coefficient $\displaystyle b$ needs to be zero.

(b)

$\displaystyle x^2+kx+(k+8)=0$

The two points will be the same if $\displaystyle b^2-4ac=0$

so you need to discover how $\displaystyle b^2-4ac\ \ne\ 0$

3. Dude I'm confused. Sorry

a. Why the coefficient of b needs to be zero?

b. So I will use > or < zero?

4. Originally Posted by hirano
Dude I'm confused. Sorry

a. Why the coefficient of b needs to be zero?

b. So I will use > or < zero?
a. Is it possible to get "roots that have equal magnitude but opposite in signs" if $\displaystyle b \neq 0$ ....? Think about it.

b. Obviously you use '> 0' (do you understand how the value of the discriminant relates to the number of solutions to a quadratic equation?)

5. a. I got it.

b. so do I need to equate (k-8)(k+4)>0?

6. Originally Posted by hirano
a. I got it.

b. so do I need to equate (k-8)(k+4)>0?
Yes, that's it.
The discriminant must be positive.