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

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

Solve for x (complex solution)

$$x=\frac{1}{2}+\sqrt{2}i\approx 0.5+1.414213562i$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

$$x-4\sqrt{x}+4=1-x$$

Subtract x+4 from both sides of the equation.

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

To find the opposite of x+4, find the opposite of each term.

$$-4\sqrt{x}=1-x-x-4$$

Combine -x and -x to get -2x.

$$-4\sqrt{x}=1-2x-4$$

Subtract 4 from 1 to get -3.

$$-4\sqrt{x}=-3-2x$$

Square both sides of the equation.

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

Expand \left(-4\sqrt{x}\right)^{2}.

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

Calculate -4 to the power of 2 and get 16.

$$16\left(\sqrt{x}\right)^{2}=\left(-3-2x\right)^{2}$$

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

$$16x=\left(-3-2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-3-2x\right)^{2}.

$$16x=9+12x+4x^{2}$$

Subtract 9 from both sides.

$$16x-9=12x+4x^{2}$$

Subtract 12x from both sides.

$$16x-9-12x=4x^{2}$$

Combine 16x and -12x to get 4x.

$$4x-9=4x^{2}$$

Subtract 4x^{2} from both sides.

$$4x-9-4x^{2}=0$$

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.

$$-4x^{2}+4x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 4 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-4±\sqrt{16+16\left(-9\right)}}{2\left(-4\right)}$$

Multiply 16 times -9.

$$x=\frac{-4±\sqrt{16-144}}{2\left(-4\right)}$$

Add 16 to -144.

$$x=\frac{-4±\sqrt{-128}}{2\left(-4\right)}$$

Take the square root of -128.

$$x=\frac{-4±8\sqrt{2}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±8\sqrt{2}i}{-8}$$

Now solve the equation x=\frac{-4±8\sqrt{2}i}{-8} when ± is plus. Add -4 to 8i\sqrt{2}.

$$x=\frac{-4+2\times 2^{\frac{5}{2}}i}{-8}$$

Divide -4+2i\times 2^{\frac{5}{2}} by -8.

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

Now solve the equation x=\frac{-4±8\sqrt{2}i}{-8} when ± is minus. Subtract 8i\sqrt{2} from -4.

$$x=\frac{-2\times 2^{\frac{5}{2}}i-4}{-8}$$

Divide -4-2i\times 2^{\frac{5}{2}} by -8.

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

The equation is now solved.

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

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

$$\sqrt{-\sqrt{2}i+\frac{1}{2}}-2=\sqrt{1-\left(-\sqrt{2}i+\frac{1}{2}\right)}$$

Simplify. The value x=-\sqrt{2}i+\frac{1}{2} does not satisfy the equation.

$$-3+\frac{1}{2}i\times 2^{\frac{1}{2}}=1+\frac{1}{2}i\times 2^{\frac{1}{2}}$$

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

$$\sqrt{\frac{1}{2}+\sqrt{2}i}-2=\sqrt{1-\left(\frac{1}{2}+\sqrt{2}i\right)}$$

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

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

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

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