MHB Real Solutions for x in a Complex Equation

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The equation x - sqrt{x^2 - x} = sqrt{x} is analyzed for real solutions with the condition x > 0. The simplification process involves dividing by sqrt{x}, leading to the conclusion that x = 1 is a solution. Substituting x = 1 back into the original equation confirms its validity, as both sides equal 1. The discussion concludes with affirmation of the solution's correctness. The focus remains on finding and verifying real solutions for the given equation.
mathdad
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Find all real solutions of the equation.

x - sqrt {x^2 - x} = sqrt {x}, where x > 0.

Since x cannot be 0, I divided through by sqrt {x} to simplify the equation.

I did a lot of math on paper and ended up with x = 1 as the answer.

I decided to substitute x = 1 into the original equation.

1 - sqrt {1^2 - 1} = sqrt {1}

1 - sqrt {1 - 1} = 1

1 - sqrt {0} = 1

1 - 0 = 1

1 = 1

What do you say?
 
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RTCNTC said:
Find all real solutions of the equation.

x - sqrt {x^2 - x} = sqrt {x}, where x > 0.

Since x cannot be 0, I divided through by sqrt {x} to simplify the equation.

I did a lot of math on paper and ended up with x = 1 as the answer.

I decided to substitute x = 1 into the original equation.

1 - sqrt {1^2 - 1} = sqrt {1}

1 - sqrt {1 - 1} = 1

1 - sqrt {0} = 1

1 - 0 = 1

1 = 1

What do you say?
Looks good.

-Dan
 
Good to know that I am right.
 
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