Solve 7+sqrt{x}=x

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

Solve for x

$$x = \frac{\sqrt{29} + 15}{2} \approx 10.192582404$$

Solution Steps

Subtract 7 from both sides of the equation.

$$\sqrt{x}=x-7$$

Square both sides of the equation.

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

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

$$x=\left(x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$x-x^{2}=-14x+49$$

Add 14x to both sides.

$$x-x^{2}+14x=49$$

Combine x and 14x to get 15x.

$$15x-x^{2}=49$$

Subtract 49 from both sides.

$$15x-x^{2}-49=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+15x-49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 15 for b, and -49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-1\right)\left(-49\right)}}{2\left(-1\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-1\right)\left(-49\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-15±\sqrt{225+4\left(-49\right)}}{2\left(-1\right)}$$

Multiply 4 times -49.

$$x=\frac{-15±\sqrt{225-196}}{2\left(-1\right)}$$

Add 225 to -196.

$$x=\frac{-15±\sqrt{29}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-15±\sqrt{29}}{-2}$$

Now solve the equation x=\frac{-15±\sqrt{29}}{-2} when ± is plus. Add -15 to \sqrt{29}.

$$x=\frac{\sqrt{29}-15}{-2}$$

Divide -15+\sqrt{29} by -2.

$$x=\frac{15-\sqrt{29}}{2}$$

Now solve the equation x=\frac{-15±\sqrt{29}}{-2} when ± is minus. Subtract \sqrt{29} from -15.

$$x=\frac{-\sqrt{29}-15}{-2}$$

Divide -15-\sqrt{29} by -2.

$$x=\frac{\sqrt{29}+15}{2}$$

The equation is now solved.

$$x=\frac{15-\sqrt{29}}{2} x=\frac{\sqrt{29}+15}{2}$$

Substitute \frac{15-\sqrt{29}}{2} for x in the equation 7+\sqrt{x}=x.

$$7+\sqrt{\frac{15-\sqrt{29}}{2}}=\frac{15-\sqrt{29}}{2}$$

Simplify. The value x=\frac{15-\sqrt{29}}{2} does not satisfy the equation.

$$\frac{13}{2}+\frac{1}{2}\times 29^{\frac{1}{2}}=\frac{15}{2}-\frac{1}{2}\times 29^{\frac{1}{2}}$$

Substitute \frac{\sqrt{29}+15}{2} for x in the equation 7+\sqrt{x}=x.

$$7+\sqrt{\frac{\sqrt{29}+15}{2}}=\frac{\sqrt{29}+15}{2}$$

Simplify. The value x=\frac{\sqrt{29}+15}{2} satisfies the equation.

$$\frac{15}{2}+\frac{1}{2}\times 29^{\frac{1}{2}}=\frac{1}{2}\times 29^{\frac{1}{2}}+\frac{15}{2}$$

Equation \sqrt{x}=x-7 has a unique solution.

$$x=\frac{\sqrt{29}+15}{2}$$