Solve left(155mright)^2=m

$${ \left(155m \right) }^{ 2 } = m$$

Solve for m

$$m=\frac{1}{24025}\approx 0.000041623$$

$$m=0$$

Steps Using Factoring

Expand \left(155m\right)^{2}.

$$155^{2}m^{2}=m$$

Calculate 155 to the power of 2 and get 24025.

$$24025m^{2}=m$$

Subtract m from both sides.

$$24025m^{2}-m=0$$

Factor out m.

$$m\left(24025m-1\right)=0$$

To find equation solutions, solve m=0 and 24025m-1=0.

$$m=0 m=\frac{1}{24025}$$

Steps Using the Quadratic Formula

Expand \left(155m\right)^{2}.

$$155^{2}m^{2}=m$$

Calculate 155 to the power of 2 and get 24025.

$$24025m^{2}=m$$

Subtract m from both sides.

$$24025m^{2}-m=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24025 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-1\right)±\sqrt{1}}{2\times 24025}$$

Take the square root of 1.

$$m=\frac{-\left(-1\right)±1}{2\times 24025}$$

The opposite of -1 is 1.

$$m=\frac{1±1}{2\times 24025}$$

Multiply 2 times 24025.

$$m=\frac{1±1}{48050}$$

Now solve the equation m=\frac{1±1}{48050} when ± is plus. Add 1 to 1.

$$m=\frac{2}{48050}$$

Reduce the fraction \frac{2}{48050} to lowest terms by extracting and canceling out 2.

$$m=\frac{1}{24025}$$

Now solve the equation m=\frac{1±1}{48050} when ± is minus. Subtract 1 from 1.

$$m=\frac{0}{48050}$$

Divide 0 by 48050.

$$m=0$$

The equation is now solved.

$$m=\frac{1}{24025} m=0$$

Steps for Completing the Square

Expand \left(155m\right)^{2}.

$$155^{2}m^{2}=m$$

Calculate 155 to the power of 2 and get 24025.

$$24025m^{2}=m$$

Subtract m from both sides.

$$24025m^{2}-m=0$$

Divide both sides by 24025.

$$\frac{24025m^{2}-m}{24025}=\frac{0}{24025}$$

Dividing by 24025 undoes the multiplication by 24025.

$$m^{2}-\frac{1}{24025}m=\frac{0}{24025}$$

Divide 0 by 24025.

$$m^{2}-\frac{1}{24025}m=0$$

Divide -\frac{1}{24025}, the coefficient of the x term, by 2 to get -\frac{1}{48050}. Then add the square of -\frac{1}{48050} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{1}{24025}m+\left(-\frac{1}{48050}\right)^{2}=\left(-\frac{1}{48050}\right)^{2}$$

Square -\frac{1}{48050} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{1}{24025}m+\frac{1}{2308802500}=\frac{1}{2308802500}$$

Factor m^{2}-\frac{1}{24025}m+\frac{1}{2308802500}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{1}{48050}\right)^{2}=\frac{1}{2308802500}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{1}{48050}\right)^{2}}=\sqrt{\frac{1}{2308802500}}$$

Simplify.

$$m-\frac{1}{48050}=\frac{1}{48050} m-\frac{1}{48050}=-\frac{1}{48050}$$

Add \frac{1}{48050} to both sides of the equation.

$$m=\frac{1}{24025} m=0$$