Solve sqrt{2-x}=x-1

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

Solve for x

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2-x}\right)^{2}=\left(x-1\right)^{2}$$

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

$$2-x=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2-x=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$2-x-x^{2}=-2x+1$$

Add 2x to both sides.

$$2-x-x^{2}+2x=1$$

Combine -x and 2x to get x.

$$2+x-x^{2}=1$$

Subtract 1 from both sides.

$$2+x-x^{2}-1=0$$

Subtract 1 from 2 to get 1.

$$1+x-x^{2}=0$$

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.

$$-x^{2}+x+1=0$$

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

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

Square 1.

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

Multiply -4 times -1.

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

Add 1 to 4.

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

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{5}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{-2} when ± is plus. Add -1 to \sqrt{5}.

$$x=\frac{\sqrt{5}-1}{-2}$$

Divide -1+\sqrt{5} by -2.

$$x=\frac{1-\sqrt{5}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from -1.

$$x=\frac{-\sqrt{5}-1}{-2}$$

Divide -1-\sqrt{5} by -2.

$$x=\frac{\sqrt{5}+1}{2}$$

The equation is now solved.

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

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

$$\sqrt{2-\frac{1-\sqrt{5}}{2}}=\frac{1-\sqrt{5}}{2}-1$$

Simplify. The value x=\frac{1-\sqrt{5}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=-\frac{1}{2}-\frac{1}{2}\times 5^{\frac{1}{2}}$$

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

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

Simplify. The value x=\frac{\sqrt{5}+1}{2} satisfies the equation.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 5^{\frac{1}{2}}\right)=\frac{1}{2}\times 5^{\frac{1}{2}}-\frac{1}{2}$$

Equation \sqrt{2-x}=x-1 has a unique solution.

$$x=\frac{\sqrt{5}+1}{2}$$