Solve x=sqrt{x+4}

$$x = \sqrt { x + 4 }$$

Solve for x

$$x = \frac{\sqrt{17} + 1}{2} \approx 2.561552813$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{x+4}\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$x^{2}=x+4$$

Subtract x from both sides.

$$x^{2}-x=4$$

Subtract 4 from both sides.

$$x^{2}-x-4=0$$

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

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

Multiply -4 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+16}}{2}$$

Add 1 to 16.

$$x=\frac{-\left(-1\right)±\sqrt{17}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{17}}{2}$$

Now solve the equation x=\frac{1±\sqrt{17}}{2} when ± is plus. Add 1 to \sqrt{17}.

$$x=\frac{\sqrt{17}+1}{2}$$

Now solve the equation x=\frac{1±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from 1.

$$x=\frac{1-\sqrt{17}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+1}{2} x=\frac{1-\sqrt{17}}{2}$$

Substitute \frac{\sqrt{17}+1}{2} for x in the equation x=\sqrt{x+4}.

$$\frac{\sqrt{17}+1}{2}=\sqrt{\frac{\sqrt{17}+1}{2}+4}$$

Simplify. The value x=\frac{\sqrt{17}+1}{2} satisfies the equation.

$$\frac{1}{2}\times 17^{\frac{1}{2}}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}\times 17^{\frac{1}{2}}$$

Substitute \frac{1-\sqrt{17}}{2} for x in the equation x=\sqrt{x+4}.

$$\frac{1-\sqrt{17}}{2}=\sqrt{\frac{1-\sqrt{17}}{2}+4}$$

Simplify. The value x=\frac{1-\sqrt{17}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}=-\left(\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}\right)$$

Equation x=\sqrt{x+4} has a unique solution.

$$x=\frac{\sqrt{17}+1}{2}$$