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

Question

Ethan Adkins · Accepted Answer

#x = O/#

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

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.

Savannah Anderson · Answer

undefined To solve the equation sqrt(x^2-28)-1=x, we can follow these steps:

1. Start by isolating the square root term on one side of the equation. Add 1 to both sides:
   sqrt(x^2-28) = x + 1

2. Square both sides of the equation to eliminate the square root:
   (sqrt(x^2-28))^2 = (x + 1)^2
   x^2 - 28 = x^2 + 2x + 1

3. Simplify the equation by canceling out the x^2 terms:
   -28 = 2x + 1

4. Move the constant term to the other side of the equation:
   2x = -29

5. Divide both sides of the equation by 2 to solve for x:
   x = -29/2

Therefore, the solution to the equation sqrt(x^2-28)-1=x is x = -29/2.