# Problem of the Week

## Updated at Nov 28, 2022 11:55 AM

This week's problem comes from the equation category.

How would you solve the equation $${(2+4(3-u))}^{2}=36$$?

Let's begin!

${(2+4(3-u))}^{2}=36$

 1 Factor out the common term $$2$$.${(2(1+2(3-u)))}^{2}=36$2 Use Multiplication Distributive Property: $${(xy)}^{a}={x}^{a}{y}^{a}$$.${2}^{2}{(1+2(3-u))}^{2}=36$3 Simplify  $${2}^{2}$$  to  $$4$$.$4{(1+2(3-u))}^{2}=36$4 Divide both sides by $$4$$.${(1+2(3-u))}^{2}=\frac{36}{4}$5 Simplify  $$\frac{36}{4}$$  to  $$9$$.${(1+2(3-u))}^{2}=9$6 Take the square root of both sides.$1+2(3-u)=\pm \sqrt{9}$7 Since $$3\times 3=9$$, the square root of $$9$$ is $$3$$.$1+2(3-u)=\pm 3$8 Break down the problem into these 2 equations.$1+2(3-u)=3$$1+2(3-u)=-3$9 Solve the 1st equation: $$1+2(3-u)=3$$.1 Subtract $$1$$ from both sides.$2(3-u)=3-1$2 Simplify  $$3-1$$  to  $$2$$.$2(3-u)=2$3 Divide both sides by $$2$$.$3-u=1$4 Subtract $$3$$ from both sides.$-u=1-3$5 Simplify  $$1-3$$  to  $$-2$$.$-u=-2$6 Multiply both sides by $$-1$$.$u=2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$u=2$10 Solve the 2nd equation: $$1+2(3-u)=-3$$.1 Subtract $$1$$ from both sides.$2(3-u)=-3-1$2 Simplify  $$-3-1$$  to  $$-4$$.$2(3-u)=-4$3 Divide both sides by $$2$$.$3-u=-\frac{4}{2}$4 Simplify  $$\frac{4}{2}$$  to  $$2$$.$3-u=-2$5 Subtract $$3$$ from both sides.$-u=-2-3$6 Simplify  $$-2-3$$  to  $$-5$$.$-u=-5$7 Multiply both sides by $$-1$$.$u=5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$u=5$11 Collect all solutions.$u=2,5$Doneu=2,5