Do Those Vertical Bars Change Anything?

Algebra Level 2

Consider the equation $\left| x \right|^2 + \left| x \right| - 6 = 0$ .

Let $n$ be the number of real roots, $S$ be the sum of those roots, and $P$ be the product of those roots. What is $\left| n + S + P \right|$ ?

The answer is 2.

2 solutions

Mathh Mathh
Jul 4, 2014

$|x|^2+|x|-6=0\iff (|x|+3)(|x|-2)=0\\\\\iff \begin{cases}|x|=-3\text{ (impossible, since }\forall a\in\mathbb R\text{, we have }|a|\ge 0 \text{)}\\\text{or}\\|x|=2\iff \begin{cases}x=2.\\\text{or}\\x=-2.\end{cases}\end{cases}$

Therefore, all the roots are $x_1=2$ and $x_2=-2$

There are $2$ of them, hence $n=2$ .

$S=x_1+x_2=2+(-2)=0$ .

$P=x_1x_2=2\cdot(-2)=-4$ .

$|n+S+P|=|2+0+(-4)|=|-2|=\boxed{2}$ .

A more simple-minded formulation: For $x < 0,$ the equation becomes $x^{2}-x-6=0$ , whose solution set is $\left\lbrace -2,3 \right\rbrace$ . But 3 is excluded. Similarly, the solution set for $x > 0$ is $\left\lbrace -3,2 \right\rbrace$ and the -3 is excluded. Then the solution set for the original equation is $\left\lbrace -2,2 \right\rbrace$

Bill Bell - 6 years, 11 months ago

Great solution, same way I did it. But I was wondering, is it possible to use vietas in this equation?

Trevor Arashiro - 6 years, 11 months ago

I guess you can....but I feel the method posted above is the best..

Krishna Ramesh - 6 years, 11 months ago

Since there is the absolute value sign , relying on negative terms as if x didn't have the absolute value sign would make vieta's in this case faulty. Using vieta's, you would had faulty term x=-3 would be there, and it wouldn't include x=-2,

Razzi Masroor - 4 years, 11 months ago

Congratulations. Nice thinking.

Niranjan Khanderia - 6 years, 10 months ago

i did same

Saksham Jain - 3 years, 6 months ago

Zenobia Roy
Aug 10, 2019

first, let's assume x is a positive number. so, the formula becomes

x*x+x=6

which can be simplify as

x(x+1)=6

since 6=2*3

then x=2

and because of absolute number means turning all the negative numbers to positive , and positive number still remaining positive ,so whether x is positive or not does not matter. so, x=2 or x=-2

then, we can get:

n=2,p=0,s=-4

This answer deserves way more upvotes

Kumudesh Ghosh - 1 year, 1 month ago

Do Those Vertical Bars Change Anything?

The answer is 2.

2 solutions

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