# Modulus functions problem sum.

1. Jul 21, 2010

### equilibrum

Hi. I've been attempting this modulus problem from my textbook for the past hour and could not find a way to get the correct answer.

1. The problem statement, all variables and given/known data
The equation x^2 -(k+2)x + 2k + 1 = 0 ,where k is a constant has two real roots α and β.
1)Express α+β and αβ in terms of k
b) if |α| = |β| , find
2)the possible values of k
3)the roots corresponding to each value of k

2. Relevant equations
α+β = -b/a
αβ = c/a
b^2-4ac = 0 / D = 0 or >/< or >= / =<

3. The attempt at a solution
Part 1 is just formula,no problem with that.

2)I presumed from the question that since the equation has two real roots(cuts at two distinct points) and/or since the modulus of alpha is equal to the modulus of beta,the roots may also be the same and therefore i applied the formula b^2 -4ac ≥ 0

(-(k+2))^2 - 4(1)(2k+1) ≥ 0
k^2-4k ≥ 0
k(k-4)≥0
By solving using quadratic inequalities,
k ≥ 4 or k ≤ 0

I referred to the answer for the possible values of k and it was 4,0,-2 . If i did a guess and check by substituting k = 4 into the equation i will get the same roots when they are in modulus form and this is also true for k = 0 . I guess this method although gives me the correct answer and proves it ,is wrong. This is because I cannot find the other possible value of k which was -2 unless i did a guess and check by going under 0 as shown in the inequality but I guess that there must be another method(as i presume that doing guess and check can yield an infinite number of solutions with decimal places etc.) . I need help on that.

Any inputs is kindly appreciated.

2. Jul 21, 2010

### ehild

Consider the cases when |α| = |β|

ehild

3. Jul 21, 2010

### Mentallic

To expand on what ehild said, modulus can be understood as being the distance from the origin. Since the distance of the two roots are equal from the origin, since $$|\alpha|=|\beta|$$ then either the roots are equal or they lie on opposite ends of the origin.

i.e. $$\alpha=\pm \beta$$

Now consider both cases, one at a time, and simplify your parabola equations for each case. For example, when the roots are of opposite signs, we have roots $$\alpha$$ and $$-\alpha$$ so we can turn the equation into the form $$(x-\alpha)(x-(-\alpha))=0$$.

4. Jul 21, 2010

### equilibrum

if we take (x-α)(x-(-α))=0 . and treat α + β as 2α we get k+2 and α = (k+2)/2 ,

(x+α)(x-α) = x^2 - α^2
after subbing α = (k+2)/2 , we finally get 4x^2 - k^2 + 4k + 4 = 0 ?
if this is correct,i don't really know how to continue either.

Last edited: Jul 22, 2010
5. Jul 22, 2010

### equilibrum

Oh oh i just tried something, wonder if this is mathematically correct,

since , |α| = |β|
By referrring to the definition of the absolute value(if a ≥ 0 and |x| = a , then x = a or x = -a)
|α| = β or |α| = -β
α+ β = 0 or α - β = 0
As we have already found α+ β to be k+2 ,
k + 2 = 0
k = -2

Since |α| = |β|,the two roots are the same and hence , graph only intersects at one point,
b^2 - 4ac = 0
-(k+2))^2 - 4(1)(2k+1) = 0
k^2-4k = 0
k(k-4) = 0
therefore, k = 0 or k = 4

6. Jul 22, 2010

### Mentallic

Yep, that's it!

But your approach in post #4 was heading there. I'll show you how you can finish it off.

We'll take the case where $$\alpha=-\beta$$ (by the way, where you said "|α| = β or |α| = -β" it should be changed to "$\alpha=\beta$ or $\alpha=-\beta$),

We have $$y=(x-\alpha)(x-\beta)$$ of course for any parabola. We should add a constant of multiplication at the front but it's not necessary here since we're trying to make it equivalent to the equation $$y=x^2-(k+2)x+2k+1$$ and the constant in front of the x2 term is just 1.

So we have $$(x-\alpha)(x-(-\alpha))=x^2-(k+2)x+2k+1$$. The constant multipliers of each term are equal, that is, in $$x^2-\alpha^2=x^2-(k+2)x+2k+1$$ the constants in front of the x2 should be equal, the constants in front of the x term are equal, so $$0=-(k+2)$$ since there is no x on the left side. And finally $$-\alpha^2=2k+1$$

Now all you can do from this is find out that k=-2 from the first equation because the second equation has 2 variables and you can't strictly solve for k. This means for the original equation to have two roots with opposite signs, only k=-2 will give that answer. Using $$(x-\alpha)^2=y$$ will give you the other two values of k as you've found already.

7. Jul 22, 2010

### equilibrum

Thanks a lot for your explanation Mentallic :)
So my method is correct as well? Except the part where I misunderstood the positive value definition.
As for your method,I understood everything you said till the part involving
$$-\alpha^2=2k+1$$
and
$$(x-\alpha)^2=y$$

How do we derive 0 and 4 from the above mentioned equations?

8. Jul 22, 2010

### Mentallic

Yes your method is correct and simple. I'm not even sure why I began to confuse you with a different method!

But anyway, if you're interested... Those two equations don't have anything to do with each other. The first was dealing with the case where the roots are of opposite signs, which means we have the equation

$$x^2-\alpha^2=0$$

and this has to be equivalent to

$$x^2 -(k+2)x + 2k + 1 = 0$$

This means the terms in front of (which is called equating the coefficients):

x2 - must be 1=1, this doesn't tell us anything interesting since it's obvious.

x - must be 0=-(k+2), this tells us k=-2

constant - must be $-\alpha^2=2k+1$, this doesn't tell us anything because we can't strictly solve for k. It's 1 equation with 2 variables.

Doing the same for the second case where the roots are equal, this means it is of the form $$(x-\alpha)^2=0$$
So then we now have the equation $$x^2-2\alpha x +\alpha^2=0$$ and now we do the same thing as we did before, equate all the coefficients. This will give us the other two values k=0,4.

9. Jul 22, 2010

### equilibrum

Dosen't hurt to experiment with alternative methods. I need to buck up in my additional mathematics in time for my 'O's next year anyway, i'm a sucker in maths :P

Anyway, I understand your method now. Comparing coefficients of like terms(as what we call it) happens a lot in the algebra i'm dealing with in my syllabus. Thank you so much for the help^^

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Modulus functions problem Date
Modulus of a complex number with hyperbolic functions Oct 21, 2017
Quadratic inequality involving Modulus Function Jan 18, 2015
Modulus function sketch Oct 30, 2013
Modulus Functions Equation Sep 24, 2010
Modulus Function Feb 13, 2010