Solve sqrt{x-1}=-x+2

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

Solve for x

$$x = \frac{5 - \sqrt{5}}{2} \approx 1.381966011$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x-1} to the power of 2 and get x-1.

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

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

$$x-1=\left(-x\right)^{2}+4\left(-x\right)+4$$

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

$$x-1=x^{2}+4\left(-x\right)+4$$

Subtract x^{2} from both sides.

$$x-1-x^{2}=4\left(-x\right)+4$$

Subtract 4\left(-x\right) from both sides.

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

Subtract 4 from both sides.

$$x-1-x^{2}-4\left(-x\right)-4=0$$

Multiply -1 and 4 to get -4.

$$x-1-x^{2}-4\left(-1\right)x-4=0$$

Multiply -4 and -1 to get 4.

$$x-1-x^{2}+4x-4=0$$

Combine x and 4x to get 5x.

$$5x-1-x^{2}-4=0$$

Subtract 4 from -1 to get -5.

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

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.

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

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-5±\sqrt{25-20}}{2\left(-1\right)}$$

Add 25 to -20.

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

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{5}}{-2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

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

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

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

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

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

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

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

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

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

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