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

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

Solve for x

$$x=3\sqrt{2}+3\approx 7.242640687$$

$$x=3-3\sqrt{2}\approx -1.242640687$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-x^{2}=x^{2}$$

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

$$6x+9=x^{2}$$

Subtract x^{2} from both sides.

$$6x+9-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-6±\sqrt{36+36}}{2\left(-1\right)}$$

Add 36 to 36.

$$x=\frac{-6±\sqrt{72}}{2\left(-1\right)}$$

Take the square root of 72.

$$x=\frac{-6±6\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±6\sqrt{2}}{-2}$$

Now solve the equation x=\frac{-6±6\sqrt{2}}{-2} when ± is plus. Add -6 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}-6}{-2}$$

Divide -6+6\sqrt{2} by -2.

$$x=3-3\sqrt{2}$$

Now solve the equation x=\frac{-6±6\sqrt{2}}{-2} when ± is minus. Subtract 6\sqrt{2} from -6.

$$x=\frac{-6\sqrt{2}-6}{-2}$$

Divide -6-6\sqrt{2} by -2.

$$x=3\sqrt{2}+3$$

The equation is now solved.

$$x=3-3\sqrt{2} x=3\sqrt{2}+3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-x^{2}=x^{2}$$

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

$$6x+9=x^{2}$$

Subtract x^{2} from both sides.

$$6x+9-x^{2}=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$6x-x^{2}=-9$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+6x=-9$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=-\frac{9}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=-\frac{9}{-1}$$

Divide -9 by -1.

$$x^{2}-6x=9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=9+9$$

Add 9 to 9.

$$x^{2}-6x+9=18$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{18}$$

Simplify.

$$x-3=3\sqrt{2} x-3=-3\sqrt{2}$$

Add 3 to both sides of the equation.

$$x=3\sqrt{2}+3 x=3-3\sqrt{2}$$