Solve sqrt{a^2-b^2}=a-b

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

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=b\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&b=0\text{ and }arg(a)<\pi \end{matrix}\right.$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$a^{2}-b^{2}=a^{2}-2ab+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}-b^{2}-a^{2}=-2ab+b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$-b^{2}=-2ab+b^{2}$$

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

$$-2ab+b^{2}=-b^{2}$$

Subtract b^{2} from both sides.

$$-2ab=-b^{2}-b^{2}$$

Combine -b^{2} and -b^{2} to get -2b^{2}.

$$-2ab=-2b^{2}$$

Cancel out -2 on both sides.

$$ab=b^{2}$$

The equation is in standard form.

$$ba=b^{2}$$

Divide both sides by b.

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

Dividing by b undoes the multiplication by b.

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

Divide b^{2} by b.

$$a=b$$

Substitute b for a in the equation \sqrt{a^{2}-b^{2}}=a-b.

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

Simplify. The value a=b satisfies the equation.

$$0=0$$

Equation \sqrt{a^{2}-b^{2}}=a-b has a unique solution.

$$a=b$$

Solve for a

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

Solution Steps

Square both sides of the equation.

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

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

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

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

$$a^{2}-b^{2}=a^{2}-2ab+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}-b^{2}-a^{2}=-2ab+b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$-b^{2}=-2ab+b^{2}$$

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

$$-2ab+b^{2}=-b^{2}$$

Subtract b^{2} from both sides.

$$-2ab=-b^{2}-b^{2}$$

Combine -b^{2} and -b^{2} to get -2b^{2}.

$$-2ab=-2b^{2}$$

Cancel out -2 on both sides.

$$ab=b^{2}$$

The equation is in standard form.

$$ba=b^{2}$$

Divide both sides by b.

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

Dividing by b undoes the multiplication by b.

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

Divide b^{2} by b.

$$a=b$$

Substitute b for a in the equation \sqrt{a^{2}-b^{2}}=a-b.

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

Simplify. The value a=b satisfies the equation.

$$0=0$$

Equation \sqrt{a^{2}-b^{2}}=a-b has a unique solution.

$$a=b$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=a\text{, }&\text{unconditionally}\\b=0\text{, }&arg(a)<\pi \end{matrix}\right.$$

Solve for b

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