Quadratic Inequalites

May 2010
4
0
Hi. First time posting, and with a big A level exam coming up tommorow - I hope to make it a good first post :)

In every single paper so far, there seems to be a Quadratic Equation (I think a Quadratci inequality question?). Heres an example of one:

"The quadratic equation (k+1)x^2 + 12x + (k-4) = 0 has real roots.

(a) Show that k^2 - 3k -40 < 0... See more
(b)Hence find the possible values of K."

If someone could walk be through both questions it would be greatly appreciated. Could meant the difference between a grade or 2!
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
Hi. First time posting, and with a big A level exam coming up tommorow - I hope to make it a good first post :)

In every single paper so far, there seems to be a Quadratic Equation (I think a Quadratci inequality question?). Heres an example of one:

"The quadratic equation (k+1)x^2 + 12x + (k-4) = 0 has real roots.

(a) Show that k^2 - 3k -40 < 0... See more
(b)Hence find the possible values of K."

If someone could walk be through both questions it would be greatly appreciated. Could meant the difference between a grade or 2!
I assumed that k is a constant

(a) Use the discriminant: \(\displaystyle \Delta = b^2-4ac\). Since you want to show there are two real roots it must be greater than 0.

I'm not too sure about why you'd show it's greater than 0, hence the above is the best bet. I would assume that since the question says there are real roots it's fine to assume part a but meh,perhaps someone else knows more?


(b) Solve the equation given in part a for k using your favourite method
 
May 2010
4
0
So, using \(\displaystyle b^2 - 4ac\) you would get \(\displaystyle 144-4(k+1)(k-4)\).

I have no idea how you get any further?
 
May 2010
22
10
for the equation (k+1)x^2 + 12x + (k-4) = 0 to have real roots, you need to have 12^2 - 4(k-4)(k+1) positive.
thus you need to have 4k^2-12k-140>=0
 
  • Like
Reactions: spongebob92

Pim

Dec 2008
91
39
The Netherlands
I am getting:

12^2 - 4(k-4)(k+1) =
144 - 4(k^2-3k-4) =
144 - 4k^2+12k+16 =
k^2-3k-40
 
May 2010
22
10
it comes from the calculation of the discriminant (b^2-4ac) which is equal to:
12^2-4(k-4)(k+1)=144-4k^2+12k+16=160-4k^2+12k (there was a mistake in my previous post. sorry)
for the equation to have real roots you need to have the above positive which is equivalent to have 40-k^2+3k positive
 
May 2010
4
0
Where about do you get the 12K in the above example?

(Thank for the continued help)
 

Pim

Dec 2008
91
39
The Netherlands
Multiplying out of brackets works as follows:

(k-4)(k+1) = k*k-4*k+1*k-4*1 = k^2-3k-4

-4(k-4)(k+1) = -4(k^2-3k-4) = -4k^2+12k+16