Author Topic: Yr 11 Parabola Question (Read 777 times) Tweet Share

Tashi · « **on:** March 30, 2010, 05:02:37 pm »

Ok this is probably an easy question but I'm stumped

Find the values of $k$ such that:

$(k-1)x^2 - 12x - (k+12) = 0$ has two solutions.

simonhu81292 · « **Reply #1 on:** March 30, 2010, 05:11:06 pm »

is the answer
k= -8 or -3?

the.watchman · « **Reply #2 on:** March 30, 2010, 05:14:03 pm »

For two solutions,

$\triangle = (-12)^2 - 4 \cdot (k-1) \cdot (-k-12) >0$

$4k^2 + 44k + 96>0$

$4(k+8)(k+3)>0$

Considering that this graph has x-ints at $k=-8$ AND $k=-3$ and is a positive graph:

Then $k<-8$ OR $k>-3$

(Try algebraic method for this one, it's neater

m@tty · « **Reply #3 on:** March 30, 2010, 05:14:34 pm »

For there to be two solutions the determinant ( $\Delta$ ) must be greater than zero.

$\Delta=b^2-4ac$

From equation $a=k-1$ , $b=-12$ and $c=-k-12$

Therefore, $\Delta=(-12)^2-4(k-1)(-k-12)$

$=144+4(k^2+11k-12)$

$=4k^2+44k+96$

For there to be two solutions this expression must be greater than zero.

$4k^2+44k+96>0$

$4(k^2+11k+24)>0$

$(k+8)(k+3)>0$

This is an upright parabola with roots at $k=-8$ and $k=-3$ .

Hence (and draw a quick sketch to confirm) $\Delta>0$ where $k\in(-\infty,-8) \cup (-3,\infty)$ (Or which I think looks cooler: $k\in \mathbb{R}\setminus[-8,-3]$ )

And for two solutions we require $\Delta>0$ So the answer is as above.

Tashi · « **Reply #4 on:** March 30, 2010, 05:26:31 pm »

Oooh ok yep thanks heaps - I get it now

I would give you karma - but I'm not up to 50 posts yet lol.

m@tty · « **Reply #5 on:** March 30, 2010, 05:29:35 pm »

No problem.

Almost 50

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Author Topic: Yr 11 Parabola Question (Read 777 times) Tweet Share

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