Determining Intersections of Quadratic Equations with a Parameter

[t_n_p]

Homework Statement

I think I've got everything down pat, I just need someone to check that I'm correct.

2 equations:

y= kx²+3x-6
y=x²-2x+3k

For what values of k do the 2 equations intersect:
a) no times
b) one time
c) 2 times

The Attempt at a Solution

equate the equations:
kx²+3x-6 = x²-2x+3k
x²(k-1) + 5x + (3k-6) = 0

det = b²-4ac where a = (k-1), b = 5, c = (3k-6)

therefore det = -12k² + 36k -1

a) for no intersection, det < 0
b) for one intersection, det = 0
c) for two intersections, det > 0

det = 0,
-12k² + 36k -1 = 0

use quadratic equation again,
k = [-9 ± √(78)]/-6

therefore

a) no intersection when k < [-9 ± √(78)]/-6
b) one intersection when k = [-9 ± √(78)]/-6
c) two intersections when k > [-9 ± √(78)]/-6


sound good?

 

[chiro]

Your reasoning is right for the questions about intersections (one - det = 0, none det < 0, two det > 0), but you have to pay attention to what you are doing when you solve for k.

Firstly the det function is wrong. My working shows the following:

det = b2-4ac where a = (k-1), b = 5, c = (3k-6)

det = b^2 - 4ac
= 25 - 4*(k-1)(3k-6)
= 25 - 4*3*(k-1)(k-2)
= 25 - 12(k-1)(k-2)
= 25 - 12(k^2 - 3k + 2)
= 25 - 12k^2 + 36k - 24
= -12k^2 + 36k + 1

So I got a "+1" term instead of a "-1" term. I'll let you figure out the critical point (ie the roots of the equation)

The other thing I want to point out is that your idea for the inequality is wrong. The graph is essentially a quadratic that is concave down (ie the curve is bounded at the top and not the bottom).

The graph will be positive when its in between the roots and it will be negative when its outside this range and not the roots themselves. If this is confusing use a package to plot the quadratic and take note when the function is positive, zero, and negative.

Like I said you got the concepts for the first part right, but not the inequality. If you're ever in doubt draw a picture to clear up any misconceptions.

 

[t_n_p]

ok think I got you.

Figured out the +1 instead of -1 no problem.

So I figured out the graph, and the intercepts are -0.0175 & 3.0275

Therefore,
one solution when det = 0 (i.e. when k = -0.0175 or 3.0275)
2 solutions when det > 0 (i.e. positive y value, i.e. -0.0175 < k < 3.0275)
no solution when det < 0 (i.e. negative y value, i.e. k < -0.0175 or k > 3.0275)

sound better now?

 

[chiro]

ok think I got you.

Figured out the +1 instead of -1 no problem.

So I figured out the graph, and the intercepts are -0.0175 & 3.0275

Therefore,
one solution when det = 0 (i.e. when k = -0.0175 or 3.0275)
2 solutions when det > 0 (i.e. positive y value, i.e. -0.0175 < k < 3.0275)
no solution when det < 0 (i.e. negative y value, i.e. k < -0.0175 or k > 3.0275)

sound better now?

Yes that looks correct well done.

 

[t_n_p]

actually spotted another mistake.

when equating the 2 equations, the constant term should be (-6-3k) not (3k-6)

 

[chiro]

actually spotted another mistake.

when equating the 2 equations, the constant term should be (-6-3k) not (3k-6)

Computational mistakes can be a common and an absolute pain in the arse. I don't know if you're allowed to, but if you suspect something is wrong, use a computer package or a website to check the answer with your paper and pen solution. I'm not saying don't do paper and pen solutions anymore, but a quick check can answer to you if you made a mistake or not.