Solve (a+c)^2=(a-c)^2

$$( a + c ) ^ { 2 } = ( a - c ) ^ { 2 }$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}+2ac+c^{2}=\left(a-c\right)^{2}$$

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

$$a^{2}+2ac+c^{2}=a^{2}-2ac+c^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+2ac+c^{2}-a^{2}=-2ac+c^{2}$$

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

$$2ac+c^{2}=-2ac+c^{2}$$

Add 2ac to both sides.

$$2ac+c^{2}+2ac=c^{2}$$

Combine 2ac and 2ac to get 4ac.

$$4ac+c^{2}=c^{2}$$

Subtract c^{2} from both sides.

$$4ac=c^{2}-c^{2}$$

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

$$4ac=0$$

The equation is in standard form.

$$4ca=0$$

Divide 0 by 4c.

$$a=0$$

Solve for c (complex solution)

$$\left\{\begin{matrix}\\c=0\text{, }&\text{unconditionally}\\c\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}+2ac+c^{2}=\left(a-c\right)^{2}$$

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

$$a^{2}+2ac+c^{2}=a^{2}-2ac+c^{2}$$

Add 2ac to both sides.

$$a^{2}+2ac+c^{2}+2ac=a^{2}+c^{2}$$

Combine 2ac and 2ac to get 4ac.

$$a^{2}+4ac+c^{2}=a^{2}+c^{2}$$

Subtract c^{2} from both sides.

$$a^{2}+4ac+c^{2}-c^{2}=a^{2}$$

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

$$a^{2}+4ac=a^{2}$$

Subtract a^{2} from both sides.

$$4ac=a^{2}-a^{2}$$

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

$$4ac=0$$

Divide 0 by 4a.

$$c=0$$

Solve for a

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}+2ac+c^{2}=\left(a-c\right)^{2}$$

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

$$a^{2}+2ac+c^{2}=a^{2}-2ac+c^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+2ac+c^{2}-a^{2}=-2ac+c^{2}$$

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

$$2ac+c^{2}=-2ac+c^{2}$$

Add 2ac to both sides.

$$2ac+c^{2}+2ac=c^{2}$$

Combine 2ac and 2ac to get 4ac.

$$4ac+c^{2}=c^{2}$$

Subtract c^{2} from both sides.

$$4ac=c^{2}-c^{2}$$

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

$$4ac=0$$

The equation is in standard form.

$$4ca=0$$

Divide 0 by 4c.

$$a=0$$

Solve for c

$$\left\{\begin{matrix}\\c=0\text{, }&\text{unconditionally}\\c\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a^{2}+2ac+c^{2}=\left(a-c\right)^{2}$$

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

$$a^{2}+2ac+c^{2}=a^{2}-2ac+c^{2}$$

Add 2ac to both sides.

$$a^{2}+2ac+c^{2}+2ac=a^{2}+c^{2}$$

Combine 2ac and 2ac to get 4ac.

$$a^{2}+4ac+c^{2}=a^{2}+c^{2}$$

Subtract c^{2} from both sides.

$$a^{2}+4ac+c^{2}-c^{2}=a^{2}$$

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

$$a^{2}+4ac=a^{2}$$

Subtract a^{2} from both sides.

$$4ac=a^{2}-a^{2}$$

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

$$4ac=0$$

Divide 0 by 4a.

$$c=0$$