Solve sqrt{m+n}=sqrt{m}+sqrt{n}

$$\sqrt { m + n } = \sqrt { m } + \sqrt { n }$$

Solve for m

$$\left\{\begin{matrix}m=0\text{, }&n\geq 0\\m\geq 0\text{, }&n=0\end{matrix}\right.$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{m+n}\right)^{2}=\left(\sqrt{m}+\sqrt{n}\right)^{2}$$

Calculate \sqrt{m+n} to the power of 2 and get m+n.

$$m+n=\left(\sqrt{m}+\sqrt{n}\right)^{2}$$

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

$$m+n=\left(\sqrt{m}\right)^{2}+2\sqrt{m}\sqrt{n}+\left(\sqrt{n}\right)^{2}$$

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

$$m+n=m+2\sqrt{m}\sqrt{n}+\left(\sqrt{n}\right)^{2}$$

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

$$m+n=m+2\sqrt{m}\sqrt{n}+n$$

Subtract m from both sides.

$$m+n-m=2\sqrt{m}\sqrt{n}+n$$

Combine m and -m to get 0.

$$n=2\sqrt{m}\sqrt{n}+n$$

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

$$2\sqrt{m}\sqrt{n}+n=n$$

Subtract n from both sides.

$$2\sqrt{m}\sqrt{n}=n-n$$

Combine n and -n to get 0.

$$2\sqrt{m}\sqrt{n}=0$$

Divide both sides by 2\sqrt{n}.

$$\frac{2\sqrt{n}\sqrt{m}}{2\sqrt{n}}=\frac{0}{2\sqrt{n}}$$

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

$$\sqrt{m}=\frac{0}{2\sqrt{n}}$$

Divide 0 by 2\sqrt{n}.

$$\sqrt{m}=0$$

Square both sides of the equation.

$$m=0$$

Substitute 0 for m in the equation \sqrt{m+n}=\sqrt{m}+\sqrt{n}.

$$\sqrt{0+n}=\sqrt{0}+\sqrt{n}$$

Simplify. The value m=0 satisfies the equation.

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

Equation \sqrt{m+n}=\sqrt{m}+\sqrt{n} has a unique solution.

$$m=0$$

Solve for n

$$\left\{\begin{matrix}n=0\text{, }&m\geq 0\\n\geq 0\text{, }&m=0\end{matrix}\right.$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{m+n}\right)^{2}=\left(\sqrt{m}+\sqrt{n}\right)^{2}$$

Calculate \sqrt{m+n} to the power of 2 and get m+n.

$$m+n=\left(\sqrt{m}+\sqrt{n}\right)^{2}$$

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

$$m+n=\left(\sqrt{m}\right)^{2}+2\sqrt{m}\sqrt{n}+\left(\sqrt{n}\right)^{2}$$

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

$$m+n=m+2\sqrt{m}\sqrt{n}+\left(\sqrt{n}\right)^{2}$$

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

$$m+n=m+2\sqrt{m}\sqrt{n}+n$$

Subtract 2\sqrt{m}\sqrt{n} from both sides.

$$m+n-2\sqrt{m}\sqrt{n}=m+n$$

Subtract n from both sides.

$$m+n-2\sqrt{m}\sqrt{n}-n=m$$

Combine n and -n to get 0.

$$m-2\sqrt{m}\sqrt{n}=m$$

Subtract m from both sides.

$$-2\sqrt{m}\sqrt{n}=m-m$$

Combine m and -m to get 0.

$$-2\sqrt{m}\sqrt{n}=0$$

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

$$\frac{\left(-2\sqrt{m}\right)\sqrt{n}}{-2\sqrt{m}}=\frac{0}{-2\sqrt{m}}$$

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

$$\sqrt{n}=\frac{0}{-2\sqrt{m}}$$

Divide 0 by -2\sqrt{m}.

$$\sqrt{n}=0$$

Square both sides of the equation.

$$n=0$$

Substitute 0 for n in the equation \sqrt{m+n}=\sqrt{m}+\sqrt{n}.

$$\sqrt{m+0}=\sqrt{m}+\sqrt{0}$$

Simplify. The value n=0 satisfies the equation.

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

Equation \sqrt{m+n}=\sqrt{m}+\sqrt{n} has a unique solution.

$$n=0$$