How do you solve #sqrt(x^2-28)-1=x#?

1 Answer
Stefan V.
Aug 5, 2015

#x = O/#

Explanation:

Start by isolating the radical on one side of the equation.

#sqrt(x^2 - 28) - color(red)(cancel(color(black)(1))) + color(red)(cancel(color(black)(1))) = x+1#

Any possible solution to this equation must satisfy the conditions

  • #x^2 - 28>=0# #-># for real numbers, you can only take the square root from positive numbers.
  • #x-1>=0 => x>=1# #-># the square root of a real number can only be a positive number.

In order for the expression under the radical to be positive, you need

#x^2 - 28 >= 0#

#x^2 >= 28#

#|x| >=sqrt(28) => x<= -sqrt(28) vv x>= sqrt(28)#

But since #x>=1# is need for the right side of the equation, your overall condition will be #x>=sqrt(28)#.

So, square both sides of the equation to get rid of the radical

#(sqrt(x^2-28))^2 = (x+1)^2#

#color(red)(cancel(color(black)(x^2))) - 28 = color(red)(cancel(color(black)(x^2))) + 2x + 1#

This is equivalent to

#2x = -29 => x = color(red)(cancel(color(black)(-29/2)))#

As you can see, this is an extraneous solution because it doesn't meet the condition #x>=sqrt(28)#. This means that the original equation has no real solutions.

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