Solve sqrt{X}=sqrt{X-5}+1

$$\sqrt { X } = \sqrt { X - 5 } + 1$$

Solve for X

$$X=9$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{X}\right)^{2}=\left(\sqrt{X-5}+1\right)^{2}$$

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

$$X=\left(\sqrt{X-5}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{X-5}+1\right)^{2}.

$$X=\left(\sqrt{X-5}\right)^{2}+2\sqrt{X-5}+1$$

Calculate \sqrt{X-5} to the power of 2 and get X-5.

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

Add -5 and 1 to get -4.

$$X=X-4+2\sqrt{X-5}$$

Subtract X from both sides.

$$X-X=-4+2\sqrt{X-5}$$

Combine X and -X to get 0.

$$0=-4+2\sqrt{X-5}$$

Swap sides so that all variable terms are on the left hand side.

$$-4+2\sqrt{X-5}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$2\sqrt{X-5}=4$$

Divide both sides by 2.

$$\sqrt{X-5}=\frac{4}{2}$$

Divide 4 by 2 to get 2.

$$\sqrt{X-5}=2$$

Square both sides of the equation.

$$X-5=4$$

Add 5 to both sides of the equation.

$$X-5-\left(-5\right)=4-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$X=4-\left(-5\right)$$

Subtract -5 from 4.

$$X=9$$

Substitute 9 for X in the equation \sqrt{X}=\sqrt{X-5}+1.

$$\sqrt{9}=\sqrt{9-5}+1$$

Simplify. The value X=9 satisfies the equation.

$$3=3$$

Equation \sqrt{X}=\sqrt{X-5}+1 has a unique solution.

$$X=9$$