Solve sqrt{x}+x-1=0

$$\sqrt { x } + x - 1 = 0$$

Solve for x

$$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

Solution Steps

Subtract x-1 from both sides of the equation.

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

To find the opposite of x-1, find the opposite of each term.

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

The opposite of -1 is 1.

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

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x-x^{2}=1$$

Subtract 1 from both sides.

$$3x-x^{2}-1=0$$

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.

$$-x^{2}+3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

$$x=\frac{-3±\sqrt{9-4}}{2\left(-1\right)}$$

Add 9 to -4.

$$x=\frac{-3±\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{5}}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{-2} when ± is plus. Add -3 to \sqrt{5}.

$$x=\frac{\sqrt{5}-3}{-2}$$

Divide -3+\sqrt{5} by -2.

$$x=\frac{3-\sqrt{5}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from -3.

$$x=\frac{-\sqrt{5}-3}{-2}$$

Divide -3-\sqrt{5} by -2.

$$x=\frac{\sqrt{5}+3}{2}$$

The equation is now solved.

$$x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$

Substitute \frac{3-\sqrt{5}}{2} for x in the equation \sqrt{x}+x-1=0.

$$\sqrt{\frac{3-\sqrt{5}}{2}}+\frac{3-\sqrt{5}}{2}-1=0$$

Simplify. The value x=\frac{3-\sqrt{5}}{2} satisfies the equation.

$$0=0$$

Substitute \frac{\sqrt{5}+3}{2} for x in the equation \sqrt{x}+x-1=0.

$$\sqrt{\frac{\sqrt{5}+3}{2}}+\frac{\sqrt{5}+3}{2}-1=0$$

Simplify. The value x=\frac{\sqrt{5}+3}{2} does not satisfy the equation.

$$1+5^{\frac{1}{2}}=0$$

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

$$x=\frac{3-\sqrt{5}}{2}$$