Solve sqrt{a+b}=sqrt{a}+sqrt{b}

$$\sqrt { a + b } = \sqrt { a } + \sqrt { b }$$

Solve for a

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

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{a+b}\right)^{2}=\left(\sqrt{a}+\sqrt{b}\right)^{2}$$

Calculate \sqrt{a+b} to the power of 2 and get a+b.

$$a+b=\left(\sqrt{a}+\sqrt{b}\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\sqrt{a}+\sqrt{b}\right)^{2}.

$$a+b=\left(\sqrt{a}\right)^{2}+2\sqrt{a}\sqrt{b}+\left(\sqrt{b}\right)^{2}$$

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

$$a+b=a+2\sqrt{a}\sqrt{b}+\left(\sqrt{b}\right)^{2}$$

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

$$a+b=a+2\sqrt{a}\sqrt{b}+b$$

Subtract a from both sides.

$$a+b-a=2\sqrt{a}\sqrt{b}+b$$

Combine a and -a to get 0.

$$b=2\sqrt{a}\sqrt{b}+b$$

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

$$2\sqrt{a}\sqrt{b}+b=b$$

Subtract b from both sides.

$$2\sqrt{a}\sqrt{b}=b-b$$

Combine b and -b to get 0.

$$2\sqrt{a}\sqrt{b}=0$$

Divide both sides by 2\sqrt{b}.

$$\frac{2\sqrt{b}\sqrt{a}}{2\sqrt{b}}=\frac{0}{2\sqrt{b}}$$

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

$$\sqrt{a}=\frac{0}{2\sqrt{b}}$$

Divide 0 by 2\sqrt{b}.

$$\sqrt{a}=0$$

Square both sides of the equation.

$$a=0$$

Substitute 0 for a in the equation \sqrt{a+b}=\sqrt{a}+\sqrt{b}.

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

Simplify. The value a=0 satisfies the equation.

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

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

$$a=0$$

Solve for b

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

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{a+b}\right)^{2}=\left(\sqrt{a}+\sqrt{b}\right)^{2}$$

Calculate \sqrt{a+b} to the power of 2 and get a+b.

$$a+b=\left(\sqrt{a}+\sqrt{b}\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\sqrt{a}+\sqrt{b}\right)^{2}.

$$a+b=\left(\sqrt{a}\right)^{2}+2\sqrt{a}\sqrt{b}+\left(\sqrt{b}\right)^{2}$$

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

$$a+b=a+2\sqrt{a}\sqrt{b}+\left(\sqrt{b}\right)^{2}$$

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

$$a+b=a+2\sqrt{a}\sqrt{b}+b$$

Subtract 2\sqrt{a}\sqrt{b} from both sides.

$$a+b-2\sqrt{a}\sqrt{b}=a+b$$

Subtract b from both sides.

$$a+b-2\sqrt{a}\sqrt{b}-b=a$$

Combine b and -b to get 0.

$$a-2\sqrt{a}\sqrt{b}=a$$

Subtract a from both sides.

$$-2\sqrt{a}\sqrt{b}=a-a$$

Combine a and -a to get 0.

$$-2\sqrt{a}\sqrt{b}=0$$

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

$$\frac{\left(-2\sqrt{a}\right)\sqrt{b}}{-2\sqrt{a}}=\frac{0}{-2\sqrt{a}}$$

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

$$\sqrt{b}=\frac{0}{-2\sqrt{a}}$$

Divide 0 by -2\sqrt{a}.

$$\sqrt{b}=0$$

Square both sides of the equation.

$$b=0$$

Substitute 0 for b in the equation \sqrt{a+b}=\sqrt{a}+\sqrt{b}.

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

Simplify. The value b=0 satisfies the equation.

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

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

$$b=0$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\neq 0\text{, }&b=0\end{matrix}\right.$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b\neq 0\text{, }&a=0\end{matrix}\right.$$