# Week 6: Solving using the Quadratic Formula

This week in Pre Calculus 11 I learned a few things that really helped me. I learned how to recognize perfect square trinomials, how to use the box method to factor, and how to use the quadratic formula when solving.

I will quickly go over each of these things.

Perfect Square Trinomials

ex. $x^2 + 10x + 25$

Always notice the first and third terms

• 1st term: $x^2$
• 2nd term: 10x
• 3rd term: 25

Notice that $x^2$ and 25 are perfect squares

$\sqrt{x^2} = x$    and    $\sqrt{25} = 5$

The 2nd term will always be $2\sqrt{1st}\cdot\sqrt{3rd}$

ex. 10x:

• $2\sqrt{x^2}\cdot\sqrt{25}$
• $2\cdot5\cdot{x}$
• 10x

This is how you recognize perfect square trinomials

Box Method (Factoring)

This box method is typically used with ugly polynomials but can be used with any expression. You want to be careful about using this method because the numbers could get big.

ex. $5x^2 + 9x + 4$

Step 1: Insert the 1st and 3rd term into the box,

Top left box: This is where you would put the first part of the expression. ex/ $5x^2$

The bottom right box: This is where the last part of your expression would go. ex/+ 4

Step 2: Multiply the 1st and 3rd term and find the products

Step 3: Chose the pair that would +. – to the 2nd term

Step 4: Insert them into the box.

* Doesn’t matter which one empty box they’re put in

*take out what is common horizontally and vertically

Step 5: Insert into the brackets and expand to check

This is how you use the box method

The quadratic formula can be used to solve any quadratic equation. This can be a really useful and quick way to solve an equation.

ex/ \$latex 3x^2 – 4x -1 = 0 %

Step 1: Determine a, b, and c and substitute into the equation.

• a = $3x^2$
• b = -4x
• c = -1

Step 3: Simplify the radicand further

Step 4: Divide by the common denominator to get it into simplest form

This is how you use the quadratic formula

Today we learned how to recognize perfect squares trinomials, how to use the box method, and how to use the quadratic formula.

# Week 5: Solving Radical Equations

This week in Pre Calculus 11, I learned how to solve radical equations. It was a little bit difficult for me at some points, because I didn’t understand squaring. Today I will teach you how to deal with these kinds of expressions.

Steps:

1. Solving
2. Restrictions
3. Checking

Example

$\sqrt{x}$ = 6

SOLVING

• remove the radical sign by squaring both sides of the equation.

$(\sqrt{x})^2$ = $6^2$

• Simplify

x = 36

Restrictions

Note: a number under a  radical sign can not be negative because a number times itself is positive weather it be -x or x. The only exception is if it has a odd power.

In the example above x needs to be a positive number so its restriction would be x > 0 or x = 0.

Check It Out

• $\sqrt{36}$ = 6
• 6 = 6

This is how to solve a basic radical equation. Now I will move on to a more difficult equation to maximize your understanding.

Example 2

1 $-3\sqrt{2x}$ = -3 $-2\sqrt{2x}$

Note: When solving eq, what you do to one side you have to do to the other.

SOLVING

1. Place the constants on side of the equation
• 1 + 3 $-3\sqrt{2x}$ = -3 + 3 $-2\sqrt{2x}$
•  4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$

Note: The radicans are the same so they can be combined together. Place them on one side of the equation.

• 4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$
• 4 $-3\sqrt{2x}$ + $3\sqrt{2x}$ = $-2\sqrt{2x}$ + $3\sqrt{2x}$
• 4 = $1\sqrt{2x}$

3. Square both sides to remove the radical sign.

• $4^2$$(\sqrt{2x})^2$
• $4^2$ = 2x
• 16 = 2x

4. Isolate x

• $\frac{16}{2}$$\frac{2x}{2}$
• $\frac{16}{2}$ = x
• 8 = x

RESTRICTIONS

x > 0   or   x = 0

Check It Out

• 1 – $3\sqrt{2(8)}$ = -3 – $2\sqrt{2(8)}$
• 1 – $3\sqrt{16}$ = -3 – $2\sqrt{16}$
• 1 – $3\cdot4$ = -3 – $2\cdot4$
• 1 – 12 = -3 – 8
• -11 = -11

This is how you solve radical equations.

Comment below if this helped you or if you have any questions. 🙂