How do you solve #\sqrt { 1} = \sqrt { ( ( - 4) - x ) ^ { 2} }#?

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Answer 1 · 2017-11-13T15:35:31

See a solution process below:

First, square each side of the equation to eliminate the radicals while keeping the equation balanced:

#(sqrt(1))^2 = (sqrt(((-4) - x)^2))^2#

#1 = ((-4) - x)^2#

#1 = (-4 - x)^2#

Squaring the right side of the equation gives:

#1 = (-4)^2 + (-4 xx -x) + (-4 xx -x) + x^2#

#1 = 16 + 4x + 4x + x^2#

#1 = 16 + 8x + x^2#

#1 = x^2 + 8x + 16#

Next, we can subtract #color(red)(1)# from each side of the equation to put the equation in standard form:

#1 - color(red)(1) = x^2 + 8x + 16 - color(red)(1)#

#0 = x^2 + 8x + 15#

#x^2 + 8x + 15 = 0#

Then, we can factor the expression on the right side of the equation as:

#(x + 3)(x + 5) = 0#

Now, we can solve each term on the right for #0# to find the solutions to the equation:

Solution 1:

#x + 3 = 0#

#x + 3 - color(red)(3) = 0 - color(red)(3)#

#x + 0 = -3#

#x = -3#

Solution 2:

#x + 5 = 0#

#x + 5 - color(red)(5) = 0 - color(red)(5)#

#x + 0 = -5#

#x = -5#

The Solutions Are:

#x = {-5, -3}#