Solving Modulus Problem: (2-x)^2 = 15 or -15

[frozen7]

( l 2-x l ^2 ) - 2 l 2-x l = 15

((2-x)^2) = 15 + 2 l 2-x l ------1
((2-x)^2) = -15 - 2 l 2-x l------2

Am I doing correctly?

 

[HallsofIvy]

( l 2-x l ^2 ) - 2 l 2-x l = 15

((2-x)^2) = 15 + 2 l 2-x l ------1
((2-x)^2) = -15 - 2 l 2-x l------2

Am I doing correctly?

How did you get 2?

I think that you intended to drop the absolute value signs.
Yes, |2-x|²l= (2-x)². Either |2-x|= (2- x) (if 2-x>= 0) or |2-x|= -(2-x)

Change the absolute values to parentheses and you will correct.

 

[frozen7]

The correct answer should be -3 and 7
From 1, I get the answer of -1,-3,5,7

Some more, how to do this question actually?

 

[Gokul43201]

You also made the mistake of flipping the sign of the constant term : you accidentally made 15 into -15.

PS : It might make things easier to first make a substitution, 2-x = y

 

[Gokul43201]

The way you solve this problem is by using the definition of the modulus function.

f(x) = |x| = x~,~~if~x \geq 0
f(x) = |x| = -x~,~~if~x < 0

Apply this definition to your problem and convert the equation to a pair of equations over two domains, one in which 2-x is negative and the other, where 2-x is non-negative.

 

[frozen7]

The first equation is correct and it is wrong for the second one. So, what should be the second equation?
Thanks.

 

[frozen7]

( ( 2-x ) ^2 ) - 2 l 2-x l = 15-----1
-( ( 2-x ) ^2 ) - 2 l 2-x l = 15----2

Do you mean this?

 

[Dr.Brain]

Divide the above moduli into two parts , one being x>2 and second being x<2 ,

Solve for both seperately and then check final value of x you get after solving under x>2 and x<2 , if it satisfies for both categories of x.

BJ