Solve sqrt{3}-sqrt{x}=sqrt{x-2}

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

Solve for x

$$x = \frac{25}{12} = 2\frac{1}{12} \approx 2.083333333$$

Solution Steps

Square both sides of the equation.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Subtract 3 from both sides.

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

Subtract 3 from -2 to get -5.

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

Divide both sides by -2\sqrt{3}.

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

Dividing by -2\sqrt{3} undoes the multiplication by -2\sqrt{3}.

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

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

$$\sqrt{x}=\frac{5\sqrt{3}}{6}$$

Square both sides of the equation.

$$x=\frac{25}{12}$$

Substitute \frac{25}{12} for x in the equation \sqrt{3}-\sqrt{x}=\sqrt{x-2}.

$$\sqrt{3}-\sqrt{\frac{25}{12}}=\sqrt{\frac{25}{12}-2}$$

Simplify. The value x=\frac{25}{12} satisfies the equation.

$$\frac{1}{6}\times 3^{\frac{1}{2}}=\frac{1}{6}\times 3^{\frac{1}{2}}$$

Equation -\sqrt{x}+\sqrt{3}=\sqrt{x-2} has a unique solution.

$$x=\frac{25}{12}$$