- #1

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( 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?

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

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

Am I doing correctly?

Last edited:

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- Thread starter frozen7
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- #1

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

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

Am I doing correctly?

Last edited:

- #2

HallsofIvy

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frozen7 said:

((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

Yes, |2-x|

Change the absolute values to parentheses and you will correct.

- #3

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From 1, I get the answer of -1,-3,5,7

Some more, how to do this question actually?

- #4

Gokul43201

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

- #5

Gokul43201

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[tex]f(x) = |x| = x~,~~if~x \geq 0 [/tex]

[tex]f(x) = |x| = -x~,~~if~x < 0 [/tex]

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.

- #6

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Thanks.

- #7

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

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

Do you mean this?

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

Do you mean this?

- #8

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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

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