+0

# Help with algebra?

0
172
2

Find all values of k so that $$x^2 - (k - 3) x - k + 6 > 0$$ for all x.

Jul 23, 2019

#1
+3

I'm how to do this but this is my best shot. So first, you can think of it as a quadratic. So simplifying the expression you get $$x^2-kx-3x-k+6>0$$

Using the quadratic formula, you get $$\frac{-(-k+3) \sqrt {(-k-3)^2-4(6-k)}}{2} > 0$$

simplify, $$\frac{k-3\sqrt{k^2+6x+9-24+4k}}{2} = \frac{k-3\sqrt{k^2+10x-13}}{2} > 0$$

I hope that helped, i just kind of bashed it out, now i hope you can solve it. I dont really know how to myself. Sorry if this post was no help at all :(

-Evan Jul 23, 2019
#2
+2

This will be a parabola  that turns upward

To find the x coordinate of the vertex we have that    (k -3) / 2

So

[(k - 3) /2]^2  - (k -3) (k - 3)/2 - k + 6 > 0

( k^2 - 6k + 9 ] / 4   - ( k^2 - 6k+ 9)/2  - k + 6 > 0      multiply through by 4

k^2 - 6k + 9  -  2(k^2 - 6k + 9)  - 4k + 24 > 0

k^2 - 6k + 9  - 2k^2 + 12k - 18 - 4k +  24 > 0

-k^2  + 2k  + 15 > 0         mutiply through by -1 and reverse the inequality sign

k^2 - 2k - 15  < 0  factor

(k - 5) ( k + 3) < 0

We have three possible intervals that solve this

(-infinity, -3)   or  ( - 3, 5)   or  ( 5, infinity)

The outside intervals do not provide the correct solution

So.....the interval   k = (-3, 5)   is correct   Jul 23, 2019