Solve x=sqrt{x+3+7}

$$x = \sqrt{ x+3+7 }$$

Solve for x

$$x = \frac{\sqrt{41} + 1}{2} \approx 3.701562119$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{x+3+7}\right)^{2}$$

Add 3 and 7 to get 10.

$$x^{2}=\left(\sqrt{x+10}\right)^{2}$$

Calculate \sqrt{x+10} to the power of 2 and get x+10.

$$x^{2}=x+10$$

Subtract x from both sides.

$$x^{2}-x=10$$

Subtract 10 from both sides.

$$x^{2}-x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$x=\frac{-\left(-1\right)±\sqrt{1+40}}{2}$$

Add 1 to 40.

$$x=\frac{-\left(-1\right)±\sqrt{41}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{41}}{2}$$

Now solve the equation x=\frac{1±\sqrt{41}}{2} when ± is plus. Add 1 to \sqrt{41}.

$$x=\frac{\sqrt{41}+1}{2}$$

Now solve the equation x=\frac{1±\sqrt{41}}{2} when ± is minus. Subtract \sqrt{41} from 1.

$$x=\frac{1-\sqrt{41}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{41}+1}{2} x=\frac{1-\sqrt{41}}{2}$$

Substitute \frac{\sqrt{41}+1}{2} for x in the equation x=\sqrt{x+3+7}.

$$\frac{\sqrt{41}+1}{2}=\sqrt{\frac{\sqrt{41}+1}{2}+3+7}$$

Simplify. The value x=\frac{\sqrt{41}+1}{2} satisfies the equation.

$$\frac{1}{2}\times 41^{\frac{1}{2}}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}\times 41^{\frac{1}{2}}$$

Substitute \frac{1-\sqrt{41}}{2} for x in the equation x=\sqrt{x+3+7}.

$$\frac{1-\sqrt{41}}{2}=\sqrt{\frac{1-\sqrt{41}}{2}+3+7}$$

Simplify. The value x=\frac{1-\sqrt{41}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}-\frac{1}{2}\times 41^{\frac{1}{2}}=-\left(\frac{1}{2}-\frac{1}{2}\times 41^{\frac{1}{2}}\right)$$

Equation x=\sqrt{x+10} has a unique solution.

$$x=\frac{\sqrt{41}+1}{2}$$