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

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

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New post Updated on: 10 Apr 2019, 14:02
2
00:00
A
B
C
D
E

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.
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Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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New post 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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Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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New post 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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Re: How many common points have y=|x| and y=|x^2-4|?  [#permalink]

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New post 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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