## The Astounding Power of Area

### 3.2 Level 0 and Level 1 Quadratics (and a taste of Level 2)

[This material appears in Section 2 of the full course on Quadratics.]

Just a few examples should do it.

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

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

The only possible difficulty here is to remember that, in algebra, positive quantities have two square roots. (See the comment in lesson 2.1.)

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

Answer: Easy. $$w=6$$ or $$w=-6$$.

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

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

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

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

EXERCISE 1: Solve:

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

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

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

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

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

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

Up a slight notch in trickiness.

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

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

$$y-4=5$$    or    $$y-4=-5$$

yielding

$$y=9$$    or   $$y=-1$$.

WARNING!  Many like to answer these questions using the $$\pm$$ symbol. But there is a potential danger as it is tempting to 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. It never hurts to take an extra two seconds for the sake of clarity.

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

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

Divide though by 4

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

to see

$$p+2=2$$   or   $$p+2=-2$$.

This yields

$$p=0$$    or    $$p=-4$$.

EXAMPLE: 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.  (But if you like to work with complex numbers …)

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

$$x+3=7$$    or    $$x+3=-7$$

yielding

$$x=4$$    or    $$x=-10$$.

EXERCISE 3: Solve.

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

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

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

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

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

EXERCISE 4:

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?

Here’s a conceptual leap!

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

Answer:  If one is extremely clever one might realize that this is a repeat of the example $$\left(x+3\right)^{2}=49$$ we just did: the quantity $$x^{2} + 6x+9=49$$ happens to equal $$(x+3)^{2}$$.

To check this, work out $$(x+3) \times (x+3)$$ as the area of a square divided into four pieces to see it is $$x^{2} + 6x+9=49$$ in disguise.

So the challenge

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

is indeed the previous example

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

in disguise. We have:

$$x+3=7$$    or    $$x+3=-7$$

$$x=4$$    or    $$x=-10$$.

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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