Solve sqrt{x}+sqrt{x-1}=sqrt{7}

$$\sqrt { x } + \sqrt { x - 1 } = \sqrt { 7 }$$

Solve for x

$$x = \frac{16}{7} = 2\frac{2}{7} \approx 2.285714286$$

Solution Steps

Subtract \sqrt{x-1} from both sides of the equation.

$$\sqrt{x}=\sqrt{7}-\sqrt{x-1}$$

Square both sides of the equation.

$$\left(\sqrt{x}\right)^{2}=\left(\sqrt{7}-\sqrt{x-1}\right)^{2}$$

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

$$x=\left(\sqrt{7}-\sqrt{x-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-\sqrt{x-1}\right)^{2}.

$$x=\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

The square of \sqrt{7} is 7.

$$x=7-2\sqrt{7}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

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

$$x=7-2\sqrt{7}\sqrt{x-1}+x-1$$

Subtract 1 from 7 to get 6.

$$x=6-2\sqrt{7}\sqrt{x-1}+x$$

Add 2\sqrt{7}\sqrt{x-1} to both sides.

$$x+2\sqrt{7}\sqrt{x-1}=6+x$$

Subtract x from both sides.

$$x+2\sqrt{7}\sqrt{x-1}-x=6$$

Combine x and -x to get 0.

$$2\sqrt{7}\sqrt{x-1}=6$$

Divide both sides by 2\sqrt{7}.

$$\frac{2\sqrt{7}\sqrt{x-1}}{2\sqrt{7}}=\frac{6}{2\sqrt{7}}$$

Dividing by 2\sqrt{7} undoes the multiplication by 2\sqrt{7}.

$$\sqrt{x-1}=\frac{6}{2\sqrt{7}}$$

Divide 6 by 2\sqrt{7}.

$$\sqrt{x-1}=\frac{3\sqrt{7}}{7}$$

Square both sides of the equation.

$$x-1=\frac{9}{7}$$

Add 1 to both sides of the equation.

$$x-1-\left(-1\right)=\frac{9}{7}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$x=\frac{9}{7}-\left(-1\right)$$

Subtract -1 from \frac{9}{7}.

$$x=\frac{16}{7}$$

Substitute \frac{16}{7} for x in the equation \sqrt{x}+\sqrt{x-1}=\sqrt{7}.

$$\sqrt{\frac{16}{7}}+\sqrt{\frac{16}{7}-1}=\sqrt{7}$$

Simplify. The value x=\frac{16}{7} satisfies the equation.

$$7^{\frac{1}{2}}=7^{\frac{1}{2}}$$

Equation \sqrt{x}=-\sqrt{x-1}+\sqrt{7} has a unique solution.

$$x=\frac{16}{7}$$