For what values of K does (x-1)^2(x+2) = K have exactly one root?

Question

as much detail as possible on the method used is appreciated. thankyou

Answer 1

How about I just give you some hints and nudges in the right direction.
Move the K to the other side: (x-1)^2(x+2) - K
This is a 3rd degree polynomial. What does the graph of a 3rd degree polynomial look like? How does the - K change the graph.? When you put a horizontal line through the graph, where does it cross the graph only once? twice? 3 times? Does finding the relative extrema help you any?

Good luck :-)
p.s. There is a solution. In fact there are infinitely many possible values for K

Answer 2

I think you may have misstated the problem. Either that or your math book misstated the problem. If what you want to know is what values of K produce exactly 1 REAL root, the answers given are not correct.

What you want is K such that the local minimum is greater than 0. Then if you consider the graph, you will see that the cubic will only cross the x-axis in one place.


So, you find the first derivative and set it equal to zero.

f(x) = x^3 - 3x + (2-K)
f`(x) = 3x^2 - 3

So, the two roots are x=-1 and x=1.

So we want K, such that, f(1) > 0
(1)^3 - 3(1) + (2-K) > 0
1-3+2-K> 0 or -K>0 or K < 0

I think that's right. Check my work!!


Actually, the other answerer has pointed out an error in my thinking. There is also exactly one root if the local maximum is less than 0, or when f(-1) < 0.

f(-1) = -1 + 3 + 2 - K < 0 => 4 - K > 0 or K > 4

Answer 3

As long as K < 0, there will be only one root. First, because the degree is odd and the leading coefficient is positive, the graph goes from the bottom, to -2, up, and then touches on 1 and then curves back upwards (because the root, 1 has multiplicity 2).

Simply sketching the graph, makes it very clear. If we were to nudge the graph up any amount, we see there is only one root, that is the one that once was at -2. So we have:

(x-1)^2(x+2) - K which means we need K < 0 in order to shift the graph upwards.

And it's not necessary that it is the cube of a polynomial because we are not considering complex roots, that is, roots of the form a+bi.

Answer 4

Expand the left hand side:

(x²-2x+1)(x+2) = K
x³ - 2x² + x + 2x² - 4x + 2 = K
x³ - 3x + 2 = K

Move K to the left:

x³-3x+2-K = 0

Now, for this equation to have _only_ one distinct root, it would have to factor as (x-r)³ = 0, where r is the root. But were this true, the equation would be of the form x³ - 3rx² + 3r²x - r³ = 0. In other words, the only way the coefficient on the x² term could be zero (as it is) is if r=0, which would mean the coefficient of the x term is also zero, and it is not. We therefore conclude that regardless of the value of K, this equation _cannot_ factor as (x-r)³ = 0, and that there is no value of k for which it has exactly one root.

Answer 5

Following on from answers to your question before about this function...

This is a cubic, so has single root below its minimum and above its maximum.

i.e. for the ranges: K < 0 and K > 4

Answer 6

Well you want K such that (x-1)^2(x+2) - K = (x+a)^3

(x-1)^2(x+2)-K=x^3-3x+2-K

I don't think there is a K that can make this work since

a^3=2-K
a^2=-3
3a=0

which doesn't work.