#### spongebob92

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
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

#### spongebob92

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?

#### papex

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

• spongebob92

#### spongebob92

papex, how do you get the 12K in the above?

#### Pim

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

#### papex

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

#### spongebob92

Where about do you get the 12K in the above example?

(Thank for the continued help)

#### Pim

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