We have to solve the equation sqrt(4-x)^2 = 5- 2x

sqrt(4-x)^2 = 5- 2x

square both the sides

=> (4 - x)^2 = (5 - 2x)^2

=> 16 + x^2 - 8x = 25 + 4x^2 - 20x

=> 3x^2 - 12x +9 = 0

=> x^2 - 4x + 3 = 0

=> x^2 - 3x - x + 3 = 0

=> x(x - 3) -1( x - 3) = 0

=> (x - 1)(x-3) = 0

For x - 1= 0, x = 1

x - 3 = 0 , x = 3

**Therefore x = 1 and 3.**

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Given the equation:

sqrt(4-x)^2 = 5- 2x

First we will simplify the square root.

We know that sqrt x^2 = l x l

==> sqrt (4-x)^2 = l 4 -x l

==> l 4 - x l = 5 -2x

Then we have two cases:

==> 4-x = 5-2x

==> x = 1

OR

-(4-x) = 5- 2x

==> -4 + x = 5-2x

==> 3x = 9

==> x = 3

Then we have two solutions:

**==> x = { 1, 3}**

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