Solve (x)=sqrt{x-5}

$$( x ) = \sqrt { x - 5 }$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{19}i}{2}\approx 0.5+2.179449472i$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{x-5}\right)^{2}$$

Calculate \sqrt{x-5} to the power of 2 and get x-5.

$$x^{2}=x-5$$

Subtract x from both sides.

$$x^{2}-x=-5$$

Add 5 to both sides.

$$x^{2}-x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-1\right)±\sqrt{1-20}}{2}$$

Add 1 to -20.

$$x=\frac{-\left(-1\right)±\sqrt{-19}}{2}$$

Take the square root of -19.

$$x=\frac{-\left(-1\right)±\sqrt{19}i}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{19}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{19}i}{2} when ± is plus. Add 1 to i\sqrt{19}.

$$x=\frac{1+\sqrt{19}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{19}i}{2} when ± is minus. Subtract i\sqrt{19} from 1.

$$x=\frac{-\sqrt{19}i+1}{2}$$

The equation is now solved.

$$x=\frac{1+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+1}{2}$$

Substitute \frac{1+\sqrt{19}i}{2} for x in the equation x=\sqrt{x-5}.

$$\frac{1+\sqrt{19}i}{2}=\sqrt{\frac{1+\sqrt{19}i}{2}-5}$$

Simplify. The value x=\frac{1+\sqrt{19}i}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}i\times 19^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}i\times 19^{\frac{1}{2}}$$

Substitute \frac{-\sqrt{19}i+1}{2} for x in the equation x=\sqrt{x-5}.

$$\frac{-\sqrt{19}i+1}{2}=\sqrt{\frac{-\sqrt{19}i+1}{2}-5}$$

Simplify. The value x=\frac{-\sqrt{19}i+1}{2} does not satisfy the equation.

$$-\frac{1}{2}i\times 19^{\frac{1}{2}}+\frac{1}{2}=-\left(\frac{1}{2}-\frac{1}{2}i\times 19^{\frac{1}{2}}\right)$$

Equation x=\sqrt{x-5} has a unique solution.

$$x=\frac{1+\sqrt{19}i}{2}$$