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25 Feb 2018, 19:14
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I'm a bit confused on how to use algebra to solve absolute value equations.
For example: If |x^2 − 12| = x, which of the following could be the value of x?
A. –4
B. –3
C. 1
D. 2
E. 3
I solved this easily using substitution, but I want to know how to solve it using algebra. I saw a similar problem and tried to copy over the steps but apparently it doesn't work algebraically here. Not sure what I am missing.
From my understanding you assume what's in the absolute value to be positive and negative, then solve for x. Assuming If |x^2 − 12| is positive, we get
x^2 - 12 = +x
x^2 -x -12 = 0
(x-4)(x+3)
x=4 or -3
Then do we just eliminate x=-3 since x has to be positive?
Next, assuming x is negative
x^2 - 12 = -x
x^2 + x - 12 = 0
(x+4)(x-3)
x=-4 or +3
Here, do we just eliminate x=+3 since we assumed x has to be negative?
Or should we not have done this step all together since we cannot assume x to be negative since |x^2 − 12| will never be a negative value?
In which case, how do we get the correct answer of E. x = 3.
For example: If |x^2 − 12| = x, which of the following could be the value of x?
A. –4
B. –3
C. 1
D. 2
E. 3
I solved this easily using substitution, but I want to know how to solve it using algebra. I saw a similar problem and tried to copy over the steps but apparently it doesn't work algebraically here. Not sure what I am missing.
From my understanding you assume what's in the absolute value to be positive and negative, then solve for x. Assuming If |x^2 − 12| is positive, we get
x^2 - 12 = +x
x^2 -x -12 = 0
(x-4)(x+3)
x=4 or -3
Then do we just eliminate x=-3 since x has to be positive?
Next, assuming x is negative
x^2 - 12 = -x
x^2 + x - 12 = 0
(x+4)(x-3)
x=-4 or +3
Here, do we just eliminate x=+3 since we assumed x has to be negative?
Or should we not have done this step all together since we cannot assume x to be negative since |x^2 − 12| will never be a negative value?
In which case, how do we get the correct answer of E. x = 3.
Archived Topic
Hi there,
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25 Feb 2018, 19:33
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jasonfodor
Firstly x is equal to modulus so x is positive..
Now you have to take modulus as POSITIVE or NEGATIVE
1) \(|x^2-12|=x.....x^2-12=x.....x^2-x-12=0....(x-4)(x+3)=0\)...
x=4 or -3
2)\(|x^2-12|=x.....-(x^2-12)=x.....x^2+x-12=0....(x+4)(x-3)=0\)...
x=-4 or 3
So two values come out 4 and 3..
4 is not given in Choices but 3 is given
So answer 3
Hope it helps
25 Feb 2018, 21:15
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jasonfodor
Discussed here: if-x-2-12-x-which-of-the-following-could-be-the-value-of-x-217526.html In case of any questions please post in that topic. Thank you.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
