### 2.2 Level Zero and Level One Quadratics

Lesson materials located below the video overview.

Alright, let’s explain that quadrangle method.

And to make this is easy as possible, let’s start with the quadratics that are ridiculously easy to solve and work our way up to the full quadrangle method.

And for fun, let’s treat this like a video game: We’ll go from level 0 to level 5.

Battle of the Squareks!

Just a few examples should do it.

EXAMPLE 11: Solve $$x^{2}=100$$.

Answer: Easy. $$x=10$$ or $$x=-10$$. □

The only possible difficulty here is that students often forget that in algebra most numbers have two square roots.

EXAMPLE 12: Solve $$w^{2}=36$$.

Answer: Easy. $$w=6$$ or $$w=-6$$. □

EXAMPLE 13: Solve $$x^{2}=17$$.

Answer: Okay, not as pretty but just as easy: $$x=\sqrt{17}$$ or $$x= -\sqrt{17}$$. □

EXAMPLE 14: Solve $$x^{2}=0$$.

Answer: Zero is the only number with just one square root: $$x=0$$ is it. □

Recall that all solutions to the practice problems appear in the COMPANION GUIDE to this QUADRATICS course.

PRACTICE 15: Solve:

a) $$x^{2}=121$$.

b) $$p^{2} = 40$$.

c) $$y^{2} + 5 = 14$$.

d) $$2x^{2} = 50$$.

e) $$x^{2}=-6$$.

PRACTICE 16: Leela was asked to solve $$x^{2}=1$$. She wrote: $$x=\sqrt 1$$ or $$x=-\sqrt 1$$. What do you think of her answer?

Here goes …

EXAMPLE 17: Solve $$(x+3)^{2}=100$$.

Answer: A tad more complicated but still easy. This problem is saying:   “Something squared is 100. ”  So the something must be $$10$$ or $$-10$$. That is:

$$x+3=10$$   or   $$x+3=-10$$

yielding:

$$x=7$$    or   $$x=-13$$.   □

EXAMPLE 18: Solve $$(y-4)^{2} = 25$$.

$$y-4=5$$    or    $$y-4=-5$$

yielding:

$$y=9$$    or   $$y=-1$$.   □

WARNING!!   Many students like to answer questions like these using the $$\pm$$ symbol. But there is a potential danger. Some will then write the following:

$$(y-4)^{2}=25$$

$$y-4=\pm 5$$

$$y=\pm 9$$

Do you see the danger? This is why I personally prefer to write things out and make sure I am thinking through everything carefully.

$$y-4=5$$    or    $$y-4 = -5$$

$$y=9$$    or   $$y=-1$$

This is now very clear and very correct!

EXAMPLE 19: Solve $$4(p+2)^{2}-16=0$$.

$$4(p+2)^{2}=16$$.

Divide though by 4:

$$(p+2)^2=4$$.

So

$$p+2=2$$   or   $$p+2=-2$$

yielding:

$$p=0$$    or    $$p=-4$$.    □

EXAMPLE 20: Solve $$(x+7)^{2}+9=0$$.

Answer: We have $$(x+7)^{2}=-9$$.   In the system of real numbers, it is impossible for a quantity squared to be negative. This equation has no solution.  □

EXAMPLE 21: Solve $$(x-1)^{2}=5$$.

$$x-1=\sqrt 5$$    or    $$x-1=-\sqrt 5$$

So:

$$x=1+\sqrt 5$$    or   $$x=1-\sqrt 5$$.   □

EXAMPLE 22: Solve $$(x+3)^{2}=49$$.

$$x+3=7$$    or    $$x+3=-7$$

yielding:

$$x=4$$    or    $$x=-10$$.   □

PRACTICE 23: Solve:

a) $$(x-1)^{2}=64$$

b) $$(p-3)^{2}=16$$

c) $$(y+1)^{2}-2=23$$

d) $$3(x-900)^{2}=300$$

e) $$(x-\sqrt 2)^{2}=5$$

PRACTICE 24:

a) How many solutions does $$(x+7)^{2}=0$$ possess? What are they?

b) How many solutions does $$(x+7)^{2}=-2$$ have? What are they?

Swarm of the Expandicons!

EXAMPLE 25: Solve $$x^{2} + 6x+9=49$$.

Answer:  If one is extremely clever one might realize that this is a repeat of example 22:

The quantity $$x^{2} + 6x+9=49$$ happens to equal $$(x+3)^{2}$$.

To check this, let’s work out $$(x+3) \times (x+3)$$ as the area of a square divided into four pieces:

Yes, we see that $$(x+3)^{2}$$ does equal $$x^{2} + 6x+9$$.  So the question:

Solve $$x^{2} + 6x+9=49$$

is indeed really the question:

Solve $$(x+3)^{2} =49$$

in disguise.

We have:

$$x+3=7$$    or    $$x+3=-7$$

$$x=4$$    or    $$x=-10$$.   □

So the challenge in level 2 is to recognize more complicated expressions as easy level 1 problems in disguise.

We’ll do this next.

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