Positive Definite Question (1 Viewer)

[frenzal_dude]

hey, i got this question, but im stuck with the last part.

For which values of k is 2x^2 + kx +18 positive definite.

b^2 - 4ac < 0 for definite.

so k^2 - 144 < 0

k^2 < 144

sqrt both sides and u get k < plus or minus 12
so k < 12 and k < -12
but those are both the same things?!?

the answer is -12<k<12

However when i draw up k^2 -144, u have x intercepts as -12 and 12, and y intercept as -144, and when its less than 0, k is between -12 and 12, so howcome u get the right answer when u draw it up, and u dont when u do it with equations?

 

[rama_v]

Thats because when you do the equation, there are always two possibilities you must consider, i.e. either
-12< k <12 OR
k<-12 or k>12

sub in a number (e.g. zero or 1) to see where the solution lies.

 

[insert-username]

If the curve is positive definite, it does not touch the x-axis, meaning that it has no real roots. For this to happen, your discriminant must be negative, thus b^2 must be less than 4ac.

a = 2
b = k
c = 18

k^2 - 4.2.18 < 0

k^2 < 144.


If k^2 is less than 144, then k must be less than +12 (a larger number gives you a product greater than 144), but greater than -12 (a smaller number again gives you a product greater than 144). Therefore, -12 < x < 12, because for any value of x between -12 and 12, b^2 is less than 4ac. The key is that k^2 - 144 is not the positive definite equation you want. Don't bother graphing it - its the solutions to that equation that are necessary.


I_F