+9676 \(\sqrt{x-1}=x\)
\(x-1=x^2\)
\(x^2-x+1=0\)
\(x=\frac{(-1)^2\pm\sqrt{4(1)(1)}}{2(1)}\)<------ Quadratic Equation
\(= \frac{1}{2}\pm\frac{2}{2}\)
x = 3/2 or x=-1/2
There are two solutions.
+33661 You should check your results Max!
You haven't done the quadratic solution correctly.
+9676 Oops. I was wrong
\(\sqrt{x-1}=x\)
\(x-1=x^2\)
\(x^2-x+1=0\)
\(x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}\)<------ Quadratic Equation
=\(\frac{1}{2}\pm\frac{\sqrt{-3}}{2}\)
=\(\frac{1\pm\sqrt3i}{2}\)
No real roots and 2 imaginary roots
Solve for x:
sqrt(x-1) = x
Raise both sides to the power of two:
x-1 = x^2
Subtract x^2 from both sides:
-x^2+x-1 = 0
Multiply both sides by -1:
x^2-x+1 = 0
Subtract 1 from both sides:
x^2-x = -1
Add 1/4 to both sides:
x^2-x+1/4 = -3/4
Write the left hand side as a square:
(x-1/2)^2 = -3/4
Take the square root of both sides:
x-1/2 = (i sqrt(3))/2 or x-1/2 = 1/2 (-i) sqrt(3)
Add 1/2 to both sides:
x = 1/2+(i sqrt(3))/2 or x-1/2 = 1/2 (-i) sqrt(3)
Add 1/2 to both sides:
Answer: | x = 1/2+(i sqrt(3))/2 or x = 1/2-(i sqrt(3))/2
