Solve sqrt{x-2}=1+sqrt{x-4}

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

Solve for x

$$x = \frac{17}{4} = 4\frac{1}{4} = 4.25$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{x-4}\right)^{2}.

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

Calculate \sqrt{x-4} to the power of 2 and get x-4.

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

Subtract 4 from 1 to get -3.

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

Subtract 2\sqrt{x-4} from both sides.

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

Subtract x from both sides.

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

Combine x and -x to get 0.

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

Add 2 to both sides.

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

Add -3 and 2 to get -1.

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

Divide both sides by -2.

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

Fraction \frac{-1}{-2} can be simplified to \frac{1}{2} by removing the negative sign from both the numerator and the denominator.

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

Square both sides of the equation.

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

Add 4 to both sides of the equation.

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

Subtracting -4 from itself leaves 0.

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

Subtract -4 from \frac{1}{4}.

$$x=\frac{17}{4}$$

Substitute \frac{17}{4} for x in the equation \sqrt{x-2}=1+\sqrt{x-4}.

$$\sqrt{\frac{17}{4}-2}=1+\sqrt{\frac{17}{4}-4}$$

Simplify. The value x=\frac{17}{4} satisfies the equation.

$$\frac{3}{2}=\frac{3}{2}$$

Equation \sqrt{x-2}=\sqrt{x-4}+1 has a unique solution.

$$x=\frac{17}{4}$$