Problem of the Week

Updated at Oct 22, 2018 8:20 AM

This week's problem comes from the equation category.

How would you solve $$(3-{m}^{2})\times \frac{3-m}{5}=\frac{44}{5}$$?

Let's begin!

$(3-{m}^{2})\times \frac{3-m}{5}=\frac{44}{5}$

1
Simplify $$(3-{m}^{2})\times \frac{3-m}{5}$$ to $$\frac{(3-{m}^{2})(3-m)}{5}$$
$\frac{(3-{m}^{2})(3-m)}{5}=\frac{44}{5}$

2
Multiply both sides by $$5$$
$(3-{m}^{2})(3-m)=44$

3
Expand
$9-3m-3{m}^{2}+{m}^{3}=44$

4
Move all terms to one side
$9-3m-3{m}^{2}+{m}^{3}-44=0$

5
Simplify $$9-3m-3{m}^{2}+{m}^{3}-44$$ to $$-35-3m-3{m}^{2}+{m}^{3}$$
$-35-3m-3{m}^{2}+{m}^{3}=0$

6
Factor $$-35-3m-3{m}^{2}+{m}^{3}$$ using Polynomial Division
$({m}^{2}+2m+7)(m-5)=0$

7
Solve for $$m$$
$m=5$

8
$m=\frac{-2+2\sqrt{6}\imath }{2},\frac{-2-2\sqrt{6}\imath }{2}$
$m=5,\frac{-2+2\sqrt{6}\imath }{2},\frac{-2-2\sqrt{6}\imath }{2}$
$m=5,-1+\sqrt{6}\imath ,-1-\sqrt{6}\imath$