Solve (m-3)^2>4

$$( m - 3 ) ^ { 2 } > 4$$

Solve for m

$$m\in \left(-\infty,1\right)\cup \left(5,\infty\right)$$

Solution Steps

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\left(m-3\right)^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula.

$$m=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$m=\frac{6±4}{2}$$

Solve the equation m=\frac{6±4}{2} when ± is plus and when ± is minus.

$$m=5 m=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-5\right)\left(m-1\right)>0$$

For the product to be positive, m-5 and m-1 have to be both negative or both positive. Consider the case when m-5 and m-1 are both negative.

$$m-5<0 m-1<0$$

The solution satisfying both inequalities is m<1.

$$m<1$$

Consider the case when m-5 and m-1 are both positive.

$$m-1>0 m-5>0$$

The solution satisfying both inequalities is m>5.

$$m>5$$

The final solution is the union of the obtained solutions.

$$m<1\text{; }m>5$$