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

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Manager
Joined: 21 Feb 2019
Posts: 105
Location: Italy
How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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Updated on: 10 Apr 2019, 14:02
2
00:00

Difficulty:

85% (hard)

Question Stats:

33% (02:16) correct 67% (01:38) wrong based on 27 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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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.
Intern
Joined: 13 Aug 2018
Posts: 18
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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10 Apr 2019, 08:33
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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Senior Manager
Joined: 28 Jul 2016
Posts: 301
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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10 Apr 2019, 11:13
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$$b^2 -4ac$$
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Manager
Joined: 21 Feb 2019
Posts: 105
Location: Italy
Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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10 Apr 2019, 12:19
3
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.
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Re: How many common points have y=|x| and y=|x^2-4|?   [#permalink] 10 Apr 2019, 12:19
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