Topic 4/5 Review (Math 2)

Topic 4/5 Review (Math 2)

Test is on Oct 20

Solving by Factoring

Always check to see if the quadratic expression can be factored easily

https://www.youtube.com/watch?v=d2lzmhEwvLo

x² + 6x + 9 = 0

Let's factor the quadratic first.

We are looking for two factors of 9 that add up to 6....

(x + 3)(x + 3) = 0

Then set the factors equal to 0 (a * b = 0; the zero-product property).

x + 3 = 0

x = -3

Now, it's your turn....

Multiple Choice

x² - 3x = 10

• 5 and 2

• 5 and -2

• -5 and 2

• -5 and -2

• No real solutions

Solving by taking the square root...

This is an excellent method if the "bx" term is missing (ax² + c = 0 or ax² = c)

Example: x² = 50

Take the square roots of both sides
x =   $\pm\sqrt{50}$   Remember that every positive # has two square roots

Then simplify the radical
x =  $\pm\sqrt{25\ \cdot2}$  
x =  $\pm5\sqrt{2}$  

Now it's your turn....

Multiple Choice

x² = 12

• $x\ =\ \pm2\sqrt{3}$

• $x\ =\ \pm3\sqrt{2}$

• $x\ =\ \pm4\sqrt{3}$

• x = 3 $\sqrt{5}$

Multiple Choice

x² = -36

• $\pm6$

• $\pm\sqrt{-6}$

• $-6$

• $\pm6i$

Simplifying Radicals

Make sure the radicand (the part under the square root symbol) contains no perfect square factors!

Example: $\sqrt{40x^2\ }\ \cdot\ \sqrt{12x^3}$  

Then factor each one...

 $\sqrt{4\cdot10\cdot x^2}\cdot\sqrt{4\cdot3\ \cdot x^2\cdot x}$  
Then remove common factors from both
 $4x^2\sqrt{30x}$  
Now it's your turn...

Multiple Choice

 $\sqrt{32x^5}\cdot\ \sqrt{24x^4}$  

•  $8x^4\sqrt{12x}$  

•  $16x^4\sqrt{3x}$  

•  $16x^2\sqrt{3x}$  

•  $8x^2\sqrt{12x}$  

Complex #s

Remember that complex #s are written in the a+bi form and
 $\sqrt{-1}\ =\ i$   and  $i^2=-1$  

 $\left(5\ -\ 3i\right)\left(2+i\right)$  
So we need to distribute    $5\left(2\ +i\right)\ -\ 3i\left(2+i\right)$  
                                              $10\ +\ 5i-6i-3i^2$   

  (simplify  $-3i^2$   & combine like terms)     $-3i^2\ =-\ 3\left(-1\right)\ =\ 3$   and our answer is...        $13-i$  
Now it's your turn...

Multiple Choice

 $\left(5+2i\right)\left(3-2i\right)$  

•  $15-4i$  

•  $19-4i$  

•  $15+4i$  

•  $15-4i-4i^2$  

Complex Conjugates

Remember that the real term is the same, but the imaginary terms are opposites

The conjugate of 5 + 3i is 5 - 3i

The conjugate of 4 - 2i is 4 + 2i

We have to use these when dividing by complex #s!

 $\frac{5}{3+i}$    

In order to divide by 3+i, we must multiply the top and bottom by the conjugate of 3 + i and then factor out any common factors

 $\frac{5}{3+i}\cdot\ \frac{3-i}{3-i}$   =  $\frac{5\left(3-i\right)}{9\ -\ i^2}$  = $\frac{5\left(3-i\right)}{9+1}$  = $\frac{5\left(3-i\right)}{10}$  =  $\frac{3-i}{2}$  

Now it's your turn....

Multiple Choice

 $(4-5i)(4+5i)$  

•  $16-25i$  

• -9

• 41

•  $16-25i^2$  

Did you notice?

When you multiply a complex # by its conjugate, you always get a positive #?

Solving by Completing the square

 $\left(\frac{b}{2}\right)^2$  

Remember: to find the C term, you use the above formula

 $x^2\ +\ 6x\ +\$  _____; to find the c term, take half of the b and square it.
So 6/2 = 3 and 3 squared is 9

Let's factor the perfect square trinomial

 $x^2\ +\ 14x\ +\ 49$  
You just need to take half of the b term!       $\left(x\ +\ 7\right)^2$   

 $x^2\ -\ 16x\ +\ 64$  is factored to  $\left(x\ -8\right)^2$  

You do need to make sure that a = 1 
Now you take a turn...

Multiple Choice

 $x^2+8x\ +\ 16\ =\ \left(\ \ \ \ \ \ \ \ \right)^2$  

• x + 8

• x - 4

• x + 4

• x - 8

Multiple Choice

 $x^2-18x\ +\ 81\ =\ \left(\ \ \ \ \ \ \ \ \right)^2$  

• x - 9

• x + 9

• x + 3

• x - 18

Now let's solve by completing the square

 $x^2\ -16x\ +\ 36\ =\ 0$  (Remember the quadratic must = 0)

First check to see if it's a PST....Is the C term  $\left(\frac{-16}{2}\right)^2$  ?        NO
So, let's move +36 by subtracting 36; this will allow us to make the PST
 $x^2-16x\ \ \ \ \ \ \ \ \ \ \ =\ -36$    Let's make the PST by adding  $\left(\frac{-16}{2}\right)^2\ =\ 64$  to both sides.   

We now have....

 $x^2\ -\ 16x\ +\ 64\ =\ -36\ +\ 64$   Factor  left side; simplify the right

 $\left(x-8\right)^2\ =\ 28$     Now take the square root of both sides
 $x-8=\pm\sqrt{28}$   and simplify the radical, if needed
 $x-8\ =\ \pm\sqrt{4\cdot7}$  
 $x-8\ =\ \pm2\sqrt{7}$   and now add 8 to both sides
 $x\ =\ 8\pm2\sqrt{7}$  , which can be rewritten as
 $x\ =\ 8\ +2\sqrt{7\ }\ and\ x\ =\ 8-2\sqrt{7}$  
Now, just like before, it's your turn...

Multiple Choice

 $x^2\ =\ 6x\ -\ 10$  (DON'T FORGET TO PUT THE EQUATION = 0)

•  $3\pm\sqrt{-1}$  

•  $-3\pm\sqrt{11}$  

•  $-3\pm i$  

•  $3\pm i$  

Solving by using the quadratic formula

 $x\ =\ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$  

Use this method when all else fails.  It actually works every time, but it can be labor intensive (a lot of work)
You substitute the a, b, and c values into the formula  $\left(ax^2\ +\ bx+c\right)$  

Multiple Choice

What's the a, b and c values in

 $3x^2-2x+1?$  

• a=3, b=2, c=1

• a=3, b=-2, c=1

 $2x^2-5x\ =\ 3$  

1) Let's first put this equation equal to zero.... $2x^2-5x-3=0$  

2) Identify a, b, and c values   
        a = 2        b = -5          c = -3
3)Then substitute them into the quadratic formula:
        $x=\frac{5\pm\sqrt{5^2-4\left(2\right)\left(-3\right)}}{2\left(2\right)}$  

 $2x^2-5x-3=0$  

 $x=\frac{5\pm\sqrt{5^2-4\left(2\right)\left(-3\right)}}{2\left(2\right)}$  

 $x=\frac{5\pm\sqrt{25-\left(-24\right)}}{4}$  

 $x\ =\ \frac{5\pm\sqrt{49}}{4}$  

 $x=\frac{5\pm7}{4}$  

 $x=\frac{5\pm7}{4}$  

So, we have  $x=\frac{5+7}{4}\ and\ \ \ x=\frac{5-7}{4}$  

 $x\ =\ 3\ and\ -\frac{1}{2}$  

Multiple Choice

Solve by using the quadratic formula:

 $2x^2+3x+8=0$  

•  $\frac{-3\pm\sqrt{71}}{4}$  

•  $\frac{-3\pm\sqrt{55}}{4}$  

•  $\frac{-3\pm i\sqrt{55}}{4}$  

•  $\frac{-3\pm5\sqrt{11}}{4}$  

The discriminant is        $b^2-4ac$   

 $b^2-4ac\ <0$  then there are 2 nonreal solutions
 $b^2-4ac\ =\ 0$   then there is only 1 real solution
 $b^2-4ac\ >\ 0$   then there are two real solutions.

For your test, be able to determine the nature of the solutions (roots).  By nature, you should be able to to determine the number of solutions and whether they are real or nonreal.