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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.
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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.
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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
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.
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.