What is the value of |x|

First of all there are no inequalities in the question.

Next, let me ask you a question: if x=4 then what does |x| equal to? |4|=4. If it were |x|=4 and we were asked to find the value of x, then yes there would be two solutions x=4 or x=-4. So, it should be the other way around.

Complete solution.

What is the value of |x|

Notice that we are asked to find the absolute value of x (|x|).

(1) |x^2 + 16| – 5 = 27 -->|x^2+16|=32 --> x^2=16 --> |x|=4. Sufficient.

(2) x^2 = 8x – 16 --> x^2-8x+16=0 --> (x-4)^2=0 --> x=4 --> |x|=4. Sufficient.

Answer: D.

Love seeing the modes and the inequalities surrounding it , i often used to be scared until one day i realised what exactly is a inequality
Something that is not equal and a new world was born but the problem was when you mix absolute values in it and that is scary ,lol

Coming to the question, I would like to share a shortcut
Whenever you have a | x + a| = c , go ahead with x+a =+/- c and solve for each possibility of positive and negative
This way , you will save yourself lot of time
But then again revisiting basics , we know that square of a number can never be negative and lets see how that evolves out in this question prompt

(1) |x^2 + 16| – 5 = 27

can be written as |x^2 + 16|= 32

(a) Positive value

using our trick for positive value x^2 + 16 = 32
So , x^2 = 16
So x^2 - 16 = 0
Voila isn't the sweet boy looking familiar ...

Let me share another concept with you ...this is a equation with degree 2 , so it will have 2 factors
x^2 = 16 doesn't mean x=4 ...it mean there are two identical factors each of which is 4
Likewise whenever you have a equation having 3 degrees , always know that it shall have 3 factors

But anyways , if both value of x is 4 , isn't |x|=4 ....
easy to understand always know that |number| is distance of that number from zero on number line ...
So quite obviously |4|=4 ..

But wait , what about our second choice

(a) Negative value

There are two reasons it wont work
firstly , x^2 is positive and secondly 16 is positive as well
So when you add two positive numbers , GMAT will punish you if you show RHS to be negative

But say you do x^2 + 16 = -32
X^2=-48
Can square of a number be positive 1^2=1 , -1^2=1 ..
In fact square and mode are those sweet little devils which hide the sign and therefore when you have them , you actually dunno what the number is
But anyways , if you have square of a number , rest assured that it's cant be negative

And so my friend , we have to reject this solution

And so pick up the positive solution and therein lies our sufficiency for this part

So , hmmmm...lets see A is alive because statement 1 is sufficient , we can bid adieu to C,E( but they are still nice fellows ..maybe we need them later but not for this question , sowwie !)


(2) Statement 2

-x^2 = 8x – 16

Rearrange it a bit and it boils down to again (x-4)^2 = 0 again two roots and each of them 4
So , mode of 4 is 4 and sweet little boy D is our answer here

pls clarify my doubt here.

statement 2 gives me the value x=4. Hence, MGMAT expln says |x| = 4 and sufficient to answer the question.

If i think inequalities as ranges in the number line, then statement 2 occupies one value x =4.

However, |x| = 4 has two values and occupies two positions +4 , -4 on the number line.
Hence, i did not choose statement 2.

Pls tell me where i went wrong in answering the q?.

Question! I can see how you (2) is sufficient. But I went about (1) in a different manner--

|x^2+16|-5=27
|x^2+16|=32
Then I broke it out into two parts...
a) x^2+16=32
x^2-16=0
(x-4)(x+4)=0
|x|=4

b) x^2+16=-32
x^2=-48... and then I kind of got stuck there.

Can you tell me where I went wrong with this logic?
Thank you!!!

Thanks, that was very helpful!

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What is the value of |x|

(1) |x^2 + 16| – 5 = 27
Case 1: |x^2 + 16| – 5 = 27
--> x^2 + 16 – 5 = 27 --> x^s = 16 --> x = 4 or -4 --> |x| = 4: sufficient
Case 2: |x^2 + 16| – 5 = 27
--> -(x^2 + 16) – 5 = 27 --> -x^2 = 6 --> x^2 = -6. No solution
(1) is sufficient

(2) x^2 = 8x – 16
(x-4)^2 = 0 --> x = 4 --> sufficient

Answer D

What is the value of |x|

(1) |x^2 + 16| – 5 = 27

(2) x^2 = 8x – 16

(1)
since the modulus is always positive

x^2 = 16
x = +/- 4 .|x| is always 4. Sufficient

(2)

x^2 -8x +16 =0
(x-4)^2 = 0, |x|=4, Sufficent

Answer is D