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Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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4 00:00

Difficulty:   85% (hard)

Question Stats: 38% (01:52) correct 63% (01:45) wrong based on 88 sessions

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How many common points have $$y=|x|$$ and $$y=|x^2 - 4|$$?

A. 0
B. 1
C. 2
D. 4
E. Infinitely many

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MEMENTO AUDERE SEMPER

Originally posted by lucajava on 10 Apr 2019, 07:12.
Last edited by lucajava on 10 Apr 2019, 14:02, edited 2 times in total.
Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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5
Algebraic way (solving the moduli case by case) is not difficult, but if you sketch the graph of the functions you will save precious time.
You just need to figure out that $$y = |x|$$ bisects the first and third quadrant for $$x ≥ 0$$ and second and fourth otherwise, meanwhile $$y = |x^2 - 4|$$ is an upward-opening parabola for $$x ≤ -2$$ and $$x ≥ 2$$, and a downward-opening one for $$-2 < x < 2$$. Count the number of intersections: 4. Et voilà, les jeux sont faits.
Attachments Scan.pdf [419.92 KiB]

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General Discussion
Intern  G
Joined: 13 Aug 2018
Posts: 36
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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lucajava wrote:
How many common points do $$y=|x|$$ and $$y=|x^2 - 4|$$ have?

A. 0
B. 1
C. 2
D. 4
E. Infinitely many

can someone explain this question please?

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Director  D
Joined: 28 Jul 2016
Posts: 600
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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1
Given y= |x|
thus y= x when x>0
y = -x when x<0
Similarly y=x^2−4 when$$|x^2−4| >0$$ or x>2 or x<-2
y = -(x^2−4) when $$|x^2−4| <0$$ or when 2>x>-2
I used the above 4 equation and solved values of x
4 such roots satisfies
using$$b^2 -4ac$$
Director  P
Joined: 24 Nov 2016
Posts: 611
Location: United States
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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lucajava wrote:
How many common points have $$y=|x|$$ and $$y=|x^2 - 4|$$?

A. 0
B. 1
C. 2
D. 4
E. Infinitely many

$$intersection(common.points):|x|=|x^2-4|$$
$$y=|x|…y=0:x≥0…y=-x:x<0$$
$$y=|x^2 - 4|…y=x^2 - 4:(x+2)(x-2)≥0=…x>2,x<-2…$$
$$y=|x^2 - 4|…y=-x^2+4:(x+2)(x-2)<0=…-2<x<2$$

$$range: ----(-2)----0----(2)----$$
$$(x>2):x=x^2-4…x^2-x-4=0…b^2-4ac=1-4(-4)=17=2.points$$
$$(0<x<2):x=-x^2+4…x^2+x-4=0…1-4(-4)=17=2.points$$
$$(-2<x<0):-x=-x^2+4…x^2-x-4=0…already.tested$$
$$(x<-2):-x=x^2-4…x^2+x-4=0…already.tested$$

$$total.points=2+2=4$$

Manager  B
Joined: 10 Dec 2017
Posts: 78
Location: India
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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4
Let's solve this question with the help of graphs

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Intern  B
Joined: 12 Aug 2019
Posts: 9
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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satya2029 wrote:
Let's solve this question with the help of graphs

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Followed the same approach. Graphing the question makes it so much easier to solve. Re: How many common points have y=|x| and y=|x^2-4|?   [#permalink] 09 Oct 2019, 12:34
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