Solve sqrt{x}+sqrt{3}=sqrt{x+7}

$$\sqrt { x } + \sqrt { 3 } = \sqrt { x + 7 }$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x}+\sqrt{3}\right)^{2}=\left(\sqrt{x+7}\right)^{2}$$

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

$$\left(\sqrt{x}\right)^{2}+2\sqrt{x}\sqrt{3}+\left(\sqrt{3}\right)^{2}=\left(\sqrt{x+7}\right)^{2}$$

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

$$x+2\sqrt{x}\sqrt{3}+\left(\sqrt{3}\right)^{2}=\left(\sqrt{x+7}\right)^{2}$$

The square of \sqrt{3} is 3.

$$x+2\sqrt{x}\sqrt{3}+3=\left(\sqrt{x+7}\right)^{2}$$

Calculate \sqrt{x+7} to the power of 2 and get x+7.

$$x+2\sqrt{x}\sqrt{3}+3=x+7$$

Subtract x from both sides.

$$x+2\sqrt{x}\sqrt{3}+3-x=7$$

Combine x and -x to get 0.

$$2\sqrt{x}\sqrt{3}+3=7$$

Subtract 3 from both sides.

$$2\sqrt{x}\sqrt{3}=7-3$$

Subtract 3 from 7 to get 4.

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

Divide both sides by 2\sqrt{3}.

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

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

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

Divide 4 by 2\sqrt{3}.

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

Square both sides of the equation.

$$x=\frac{4}{3}$$

Substitute \frac{4}{3} for x in the equation \sqrt{x}+\sqrt{3}=\sqrt{x+7}.

$$\sqrt{\frac{4}{3}}+\sqrt{3}=\sqrt{\frac{4}{3}+7}$$

Simplify. The value x=\frac{4}{3} satisfies the equation.

$$\frac{5}{3}\times 3^{\frac{1}{2}}=\frac{5}{3}\times 3^{\frac{1}{2}}$$

Equation \sqrt{x}+\sqrt{3}=\sqrt{x+7} has a unique solution.

$$x=\frac{4}{3}$$