The graphs y = x^2 + 4 and y = |x| + 4 have how many points in common?

Question

The graphs  y = x^2 + 4  and  y = |x| + 4  have how many points in common?

A. none
B. one
C. two
D. three
E. four

KarishmaB · Accepted Answer

If y = 0,

x^2 = 4
means x = 2 or -2

|x| = 4
means x is 4 or -4

There is no common point when y = 0.

Also, you are ignoring any points that may be common but do not lie on either the X axis or the Y axis.

This is our question: For what values of x will y be the same? 
 x^2 + 4 = |x| + 4 
 |x|^2 - |x| =0

(because  x^2 = |x|^2 )

|x| (|x| - 1) = 0

So either |x| = 0 which means x = 0
or |x| = 1 which means x = 1 or -1

So at these three points x = 0, 1, -1, values of y will be the same for the two equations.

Answer (D)

yashikaaggarwal · Answer

To know about the common points of these two lines we have to put them equal.
Such that,
X^2 + 4 = |X| + 4
X^2 = |X|
Now,
Only 3 values of X satisfy this.
-1,0,&1
So, there will be 3 points, which coincide with other line.

Answer is D

Posted from my mobile device

gsw2021 · Answer

Hi, so the way I worked it out was...

If x = 0 then for both the equations y = 4 so that is a common point.

If y = 0 then, for the first equation  x^2 = 4 
For the second equation |x| = 4 or -4

So the Positive 4 becomes the second common point. 
Total 2 common points - Answer choice C

Please explain how this approach is wrong. 
 Bunuel   VeritasKarishma

sallyvn · Answer

x =-1 / 0 / 1 => 3 common points =>D

Anamika1496 · Answer

Step 1 : Equate both the equation

X2 + 4 = |x| + 4

Step 2 : Square both sides

Result : x4 + 16 + 8x2 = x2 + 16 + 8x

Step 3 : rearrange to get final equation

Result : x(x3 + 7x - 8) = 0

OR

X3 + 7x - 8 = 0

=> this is an equation of degree 3, hence 3 solutions

Ans. D

Attachment:

0E6ACC86-018B-4F60-85F2-EB53E4935A72.jpeg [ 3.29 MiB | Viewed 5795 times ]

BrushMyQuant · Answer

y = x^2 + 4  and  y = |x| + 4

=>   x^2 + 4 = |x| + 4  
=>  x^2 = |x|

As we have |x| in the equation so we will have two cases 
 
   -   Case 1:  x ≥ 0
 
=> |x| = x
=>  x^2 = x 
=>  x^2 - x = 0 
=> x*(x-1) = 0
=> x = 0, 1

But condition was x ≥ 0 and both 0 and 1 are ≥ 0 
=>  x = 0, 1 are  SOLUTIONS   
  -   Case 2:  x ≤ 0
 
=> |x| = -x
=>  x^2 = -x 
 =>  x^2 + x = 0 
=> x*(x+1) = 0
=> x = 0, -1

But condition was x ≤ 0 and both 0 and 1 are ≤ 0 
=>  x = 0, -1 are  SOLUTIONS

=> There are 3 points in common
=> When x = -1, y = (-1)^2 + 4 = 5 
=> When x = 0, y = 0^2 + 4 = 4
=> When x = 1, y = 1^2 + 4 = 5
=> Three points are (-1,5), (0,4), (1,5)

So,  Answer will be D 
Hope it helps!

Watch the following video to MASTER Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

bumpbot · Answer

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The graphs y = x^2 + 4 and y = |x| + 4 have how many points in common?

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As we have \|x\| in the equation so we will have two cases
-Case 1: x ≥ 0 => \|x\| = x => \(x^2 = x\) => \(x^2 - x = 0\) => x*(x-1) = 0 => x = 0, 1 But condition was x ≥ 0 and both 0 and 1 are ≥ 0 => x = 0, 1 are SOLUTIONS	-Case 2: x ≤ 0 => \|x\| = -x => \(x^2 = -x\) => \(x^2 + x = 0\) => x*(x+1) = 0 => x = 0, -1 But condition was x ≤ 0 and both 0 and 1 are ≤ 0 => x = 0, -1 are SOLUTIONS

GMAT Problem Solving (PS) Questions

The graphs y = x^2 + 4 and y = |x| + 4 have how many points in common?

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