Help me find k

**hirano** · October 9th 2010, 01:54 AM

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.

**Archie Meade** · October 9th 2010, 02:12 AM

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.

Use the quadratic formula.

(a)

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

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

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

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

(b)

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

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

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

**hirano** · October 9th 2010, 02:30 AM

Dude I'm confused. Sorry

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

b. So I will use > or < zero?

**mr fantastic** · October 9th 2010, 04:06 AM

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 $b \neq 0$ ....? 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?)

**hirano** · October 9th 2010, 04:36 AM

a. I got it.

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

**Archie Meade** · October 9th 2010, 10:58 AM

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.

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