Thread: Quadratics #3

Hi

For the question:

Show that the equation (x^2 + 5x)/(7x + 9) = (1 - k)/(1 + k) has real roots for all values of k.

I know that for real roots, the discriminant has to be greater or equal than 0. I found out the discriminant, which is x^2 + kx^2 - 2x + 12 xk - 9k + 9 greater than or equal to 0. How do I make it into a quadratic equation?

Thanx

Hello,

Originally Posted by xwrathbringerx

Hi

For the question:

Show that the equation (x^2 + 5x)/(7x + 9) = (1 - k)/(1 + k) has real roots for all values of k.

I know that for real roots, the discriminant has to be greater or equal than 0. I found out the discriminant, which is x^2 + kx^2 - 2x + 12 xk - 9k + 9 greater than or equal to 0. How do I make it into a quadratic equation?

Thanx

No, what you got is just the equation from which you want to find the discriminant.

It's x²+kx²-2x+12kx+9k-9=0

which can be grouped this way :
(1+k)x²+x(12k-2)+(9k-9)=0

you know that the discriminant of ax²+bx+c is b²-4ac.
Here, a=1+k, b=12k-2, c=9k-9.


note that , that is to say


and you'll have to state the case for , that is to say

Hmm I keep on getting 27k^2 - 12k + 10 as the discriminant but when I try to find k, there are no real solutions. What does that mean?

Originally Posted by xwrathbringerx

Hmm I keep on getting 4(27k^2 - 12k + 10) as the discriminant but when I try to find k, there are no real solutions. What does that mean?

Good work, the discriminant is correct =)

Now, you want the discriminant to be positive. Once again, find the determinant of the quadratic 27k²-12k+10 :
- if it has no real root (that is to say the discriminant is < 0), then it is always positive, which is what you want
- if the discriminant is = 0, then it is positive (as a square)
- if the discriminant is > 0, it can be either positive or negative (in general)