Solve m^2+(3sqrt{3}m)^2=(3-m)^2

$$m ^ { 2 } + ( 3 \sqrt { 3 } m ) ^ { 2 } = ( 3 - m ) ^ { 2 }$$

Solve for m

$$m=\frac{2\sqrt{7}-1}{9}\approx 0.476833625$$

$$m=\frac{-2\sqrt{7}-1}{9}\approx -0.699055847$$

Steps Using the Quadratic Formula

Expand \left(3\sqrt{3}m\right)^{2}.

$$m^{2}+3^{2}\left(\sqrt{3}\right)^{2}m^{2}=\left(3-m\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$m^{2}+9\left(\sqrt{3}\right)^{2}m^{2}=\left(3-m\right)^{2}$$

The square of \sqrt{3} is 3.

$$m^{2}+9\times 3m^{2}=\left(3-m\right)^{2}$$

Multiply 9 and 3 to get 27.

$$m^{2}+27m^{2}=\left(3-m\right)^{2}$$

Combine m^{2} and 27m^{2} to get 28m^{2}.

$$28m^{2}=\left(3-m\right)^{2}$$

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

$$28m^{2}=9-6m+m^{2}$$

Subtract 9 from both sides.

$$28m^{2}-9=-6m+m^{2}$$

Add 6m to both sides.

$$28m^{2}-9+6m=m^{2}$$

Subtract m^{2} from both sides.

$$28m^{2}-9+6m-m^{2}=0$$

Combine 28m^{2} and -m^{2} to get 27m^{2}.

$$27m^{2}-9+6m=0$$

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.

$$27m^{2}+6m-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 27 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-6±\sqrt{6^{2}-4\times 27\left(-9\right)}}{2\times 27}$$

Square 6.

$$m=\frac{-6±\sqrt{36-4\times 27\left(-9\right)}}{2\times 27}$$

Multiply -4 times 27.

$$m=\frac{-6±\sqrt{36-108\left(-9\right)}}{2\times 27}$$

Multiply -108 times -9.

$$m=\frac{-6±\sqrt{36+972}}{2\times 27}$$

Add 36 to 972.

$$m=\frac{-6±\sqrt{1008}}{2\times 27}$$

Take the square root of 1008.

$$m=\frac{-6±12\sqrt{7}}{2\times 27}$$

Multiply 2 times 27.

$$m=\frac{-6±12\sqrt{7}}{54}$$

Now solve the equation m=\frac{-6±12\sqrt{7}}{54} when ± is plus. Add -6 to 12\sqrt{7}.

$$m=\frac{12\sqrt{7}-6}{54}$$

Divide -6+12\sqrt{7} by 54.

$$m=\frac{2\sqrt{7}-1}{9}$$

Now solve the equation m=\frac{-6±12\sqrt{7}}{54} when ± is minus. Subtract 12\sqrt{7} from -6.

$$m=\frac{-12\sqrt{7}-6}{54}$$

Divide -6-12\sqrt{7} by 54.

$$m=\frac{-2\sqrt{7}-1}{9}$$

The equation is now solved.

$$m=\frac{2\sqrt{7}-1}{9} m=\frac{-2\sqrt{7}-1}{9}$$

Steps for Completing the Square

Expand \left(3\sqrt{3}m\right)^{2}.

$$m^{2}+3^{2}\left(\sqrt{3}\right)^{2}m^{2}=\left(3-m\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$m^{2}+9\left(\sqrt{3}\right)^{2}m^{2}=\left(3-m\right)^{2}$$

The square of \sqrt{3} is 3.

$$m^{2}+9\times 3m^{2}=\left(3-m\right)^{2}$$

Multiply 9 and 3 to get 27.

$$m^{2}+27m^{2}=\left(3-m\right)^{2}$$

Combine m^{2} and 27m^{2} to get 28m^{2}.

$$28m^{2}=\left(3-m\right)^{2}$$

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

$$28m^{2}=9-6m+m^{2}$$

Add 6m to both sides.

$$28m^{2}+6m=9+m^{2}$$

Subtract m^{2} from both sides.

$$28m^{2}+6m-m^{2}=9$$

Combine 28m^{2} and -m^{2} to get 27m^{2}.

$$27m^{2}+6m=9$$

Divide both sides by 27.

$$\frac{27m^{2}+6m}{27}=\frac{9}{27}$$

Dividing by 27 undoes the multiplication by 27.

$$m^{2}+\frac{6}{27}m=\frac{9}{27}$$

Reduce the fraction \frac{6}{27} to lowest terms by extracting and canceling out 3.

$$m^{2}+\frac{2}{9}m=\frac{9}{27}$$

Reduce the fraction \frac{9}{27} to lowest terms by extracting and canceling out 9.

$$m^{2}+\frac{2}{9}m=\frac{1}{3}$$

Divide \frac{2}{9}, the coefficient of the x term, by 2 to get \frac{1}{9}. Then add the square of \frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+\frac{2}{9}m+\left(\frac{1}{9}\right)^{2}=\frac{1}{3}+\left(\frac{1}{9}\right)^{2}$$

Square \frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+\frac{2}{9}m+\frac{1}{81}=\frac{1}{3}+\frac{1}{81}$$

Add \frac{1}{3} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}+\frac{2}{9}m+\frac{1}{81}=\frac{28}{81}$$

Factor m^{2}+\frac{2}{9}m+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m+\frac{1}{9}\right)^{2}=\frac{28}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{1}{9}\right)^{2}}=\sqrt{\frac{28}{81}}$$

Simplify.

$$m+\frac{1}{9}=\frac{2\sqrt{7}}{9} m+\frac{1}{9}=-\frac{2\sqrt{7}}{9}$$

Subtract \frac{1}{9} from both sides of the equation.

$$m=\frac{2\sqrt{7}-1}{9} m=\frac{-2\sqrt{7}-1}{9}$$