Maths help?

Question

For what values of k will the expression x^2 - (k+2)x + (3x+6) be positive definite?!

I have a test coming up. . . .
and i Need all the help I can get.

Thanks =]

Answer 1

Using the form ax² + bx + c.....

For a definite quadratic, the discriminant (b² - 4ac) must be negative so that there are no real roots. This is taken from the quadratic formula, where it appears under the square root symbol, so if it is negative, it has no square roots and thus there are no real solutions. For this to be positive definite, a must be positive too (which it already is). So if both of these conditions are true then you have a positive definite expression.
a > 0 (we know this)
b² - 4ac < 0.

Now in the expression,
a = 1
b = - (k + 2)
c = 3x + 6

Therefore b² - 4ac = [-(k + 2)]² - 4*1*(3x + 6)
= (k + 2)² - 12x - 24
= k² + 4k + 4 - 12x - 24
= k² - 8k - 20.

So if b² - 4ac < 0, then k² - 8k - 20 < 0
Factorise this
(k - 10)(k + 2) < 0
Consider (k - 10)(k + 2) = 0 and sketch a simple graph of this along the x-axis. You will have an upwards-facing parabola with roots at -2 and 10, and you can see that (k - 10)(k + 2) < 0 is true for -2 < k < 10.

Answer 2

x^2 - (k+2)x + (3x+6)
= x^2 -kx+2x+3x + 6
= x^2 +(-k+5)x + 6
If k = 5, the expression will be positive for all values of x