## 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.]

Let’s now explain the *quadatus* method for solving quadratic equations. We’ll start with quadratic equations that are overtly connected to squares and tweak our sophistication from there.

**LEVEL 0 QUADRATIC EQUATIONS**

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?*

**LEVEL 1 QUADRATIC EQUATIONS**

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* \((y-4)^{2} = 25\).

**Answer**: We have:

\(y-4=5\) 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* \(4(p+2)^{2}-16=0\).

**Answer**: Add 16

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

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* \((x-1)^{2}=5\).

**Answer**: We have

\(x-1=\sqrt 5\) or \(x-1=-\sqrt 5\)

so

\(x=1+\sqrt 5\) or \(x=1-\sqrt 5\).

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

**Answer**: We have

\(x+3=7\) 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?*

**LEVEL 2 QUADRATICS**

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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