Solve (a-x)^2+3^2=x^2

$$( a - x ) ^ { 2 } + 3 ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=\frac{a}{2}+\frac{9}{2a}$$

$$a\neq 0$$

Steps for Solving Linear Equation

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

$$a^{2}-2ax+x^{2}+3^{2}=x^{2}$$

Calculate 3 to the power of 2 and get 9.

$$a^{2}-2ax+x^{2}+9=x^{2}$$

Subtract x^{2} from both sides.

$$a^{2}-2ax+x^{2}+9-x^{2}=0$$

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

$$a^{2}-2ax+9=0$$

Subtract a^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2ax+9=-a^{2}$$

Subtract 9 from both sides.

$$-2ax=-a^{2}-9$$

The equation is in standard form.

$$\left(-2a\right)x=-a^{2}-9$$

Divide both sides by -2a.

$$\frac{\left(-2a\right)x}{-2a}=\frac{-a^{2}-9}{-2a}$$

Dividing by -2a undoes the multiplication by -2a.

$$x=\frac{-a^{2}-9}{-2a}$$

Divide -a^{2}-9 by -2a.

$$x=\frac{a}{2}+\frac{9}{2a}$$

Solve for a (complex solution)

$$a=-\sqrt{x^{2}-9}+x$$

$$a=\sqrt{x^{2}-9}+x$$

Solve for a

$$a=-\sqrt{x^{2}-9}+x$$

$$a=\sqrt{x^{2}-9}+x\text{, }|x|\geq 3$$