## Finding Solutions

Using Factors:

1. 8x^2-18x-5=0

x=-.25 (-1/4) x=2.5 (5/2)

(4x+1)(5x-2)

- To find the factors you enter the quadratic formula into the calculator. Find both zeros, which for this equations is (-1/4) and (5/2). To turn zeros into factors you need to get both the x and the zero on the same side. For fractions the denominator or the bottom number to moved in front of the x, (4x+1) and (2x-5). Don't forget to switch the sign of the zeros when turned into factors.

2. x^2+x=12

-12 -12

x^2+x-12=0

x=3 x=-4

-3 +4

(x-3)(x+4)

- To find the factors for this one you again enter it in to the in the y= of the calculator. Find both zeros, which would be x=3 and x=-4. To get the factors you need to get the x and the zero on both sides, so subtract 3 for x=3 and add four for x=-4. Therefore the factors you get are (x-3) and (x+4).

Using Square Root Method:

Real: x^2-1=24

+1 +1

x^2=25

√x^2=√25

x=5

-We need to get the numbers on one side and the x on the other. So add one to both sides. Now that both the numbers are on one side and the x is on the other we need to get rid of the squared. So square the x^2, and what you do to one side you need to do to the other, so square root 25 too. You end up with x=5.

Imaginary: 3x^2+28=1

-28 -28

3x^2=-27

/3 /3

x^2=-9

√x^2=√-9

x=3i

-We need to get the numbers on one side and the x on the other, so we subtract 28 from both sides. Now we need to get the 3 over, the opposite of multiplication is division. So we divide both sides by 3. Then square root both sides. The number inside the square root is a negative, so when we look for pairs to come out of the square root an i also comes out. The i comes out in place of the negative. We end of with a pair of 3's, so a 3 and an i come out. Our ending answer is x=+-3i.

Real: x^2-5x-6

(-5)^2-4(1)(-6)=49

5+-√49

/2

5+-7

/2

6,-1

-This problem we will use the discriminant so remember the standard ax^2+bx+c=0. To find the discriminant we use the formula b^2-4(a)(c). So square the -5, but remember to put parenthesis around -5. The discriminant is 49, so now we will put it in -b+-√b^2-4ac

/2. The b value becomes positive since it is already negative. we know the discriminant is 49, so it goes into the square root. Then we multiply the a value by 2. When we square root the 49, we get a pair of sevens, so a seven comes out. Now add 5 plus 7 divided by 2, and 5 minus 7 divided 2. 6 and -1 are our answers.

Imaginary:

x^2+2x+2

(2)^2-4(1)(2)=-4

-2+-√-4

/2

-2+-2i

/2

-1+-i

-Find the discriminant, using (b)^2-4ac, which we find to be -4. So when we square root -4, we get a pair of two's and an i for the negative. So the equation becomes -2+-2i

/2. We can divide the two out because they are all divisible by 2. So we should end up with -1+-i.

Completing the Square Root

Real:-2x^2+4x+6

/-2

x^2-2x-3=0

+3 +3

x^2-2x+1=3+1

(x-1)^2=4

√(x-1)^2=√4

x-1=+-√4

x-1=+-2

+1 +1

x=1+-2

x=3, -1

-The first thing we do is divide everything by -2. Then set the equation equal to zero. Add 3 to both sides to get on the right side of the equal sign. Divide 2 by 2 and square it, to get 1. So 1 is what we add to both sides. we pull the x down and the first sign, which is a negative. The square root of 1 is 1, so we get (x-1)^2=4. Now we square root both sides, so the squared sign goes away and becomes x-1=+-√4. We get a pair of two's, so a 2 comes out. Now add the 1 to both sides to get x=1+-2. We now do 1+2, and 1-2, to get our answers 3 and -1.

Imaginary: x^2+6x+14=0

-14 -14

x^2+6x+9=-14+9

(x+3)^2=-5

√(x+3)^2=√-5

x+3=+-√-5

-3 -3

x=-3+-√-5

x=-3+-i√5

-There is not a number in front of the x, so we start by setting the equation equal to zero and subtracting 14 from both sides. divide 6 by 2 and square it, and we find we are adding 9 to both sides. Now square root both sides to get rid of the square. Subtract 3 from both sides to get it on the right side. It goes in front of the +- signs in front of the square root. So an i comes out, but 5 can't because it can not be broken down any smaller so it stays in the square root and we end up with x=-3+-i√5.

## Discriminant- (b)^2-4(a)(c)

One Real Solution: 9x^2+30x+25

• Set it up like the formula, (30)^2-4(9)(25)=0

-The zero means that it only touches the x-axis once.

-If the equation was entered into the y= in the calculator you would see that the graph would only touch once.

Two Real Solution: 4x^2-6x-7

• Set it up like the formula , (-6)^2-4(4)(-7)=148

-The number means that it touches the x-axis twice.

Two Imaginary Solution: 2x^2+6x+7

• Set it up like the formula, (6)^2-4(2)(7)=-20

-The negative number means that the graph doesn't touch the x-axis at all

## Turnig Standard Form into Vertex Form

x^2+6x+13=0

-13 -13

x^2+6x+9=-13=9

(x+3)^2=-4

+4 +4

f(x)=(x+3)^2+4

-Since there is no number in front of x, we will start by setting the equation equal to zero. Subtract 13 from both sides. Now divide the 6 by 2 and square it to get 9, and that is what we will add to both sides. Add 4 to both sides and we should end up with a vertex equation f(x)=(x+3)^2+4

-3x^2 +12x+3

/-3

x^2-4x-1=0

+1 +1

x^2-4x+4=1+4

(x-2)^2=5

-5 -5

f(x)=-3(x-2)^2-5

-Since there is a number in front of the x we will divide everything by -3. Now we set the equation equal to zero and add 1 to both sides. We will subtract the 5 from both sides. The -3 that we divided everything by, is brought back into the vertex form as the d value which is in front of the equation. The vertex form we get is f(x)=-3(x-2)^2-5.

## Transformations

Vertical Shift: (x-1)^2+4, (x-1)^2-4

- The vertical shift is the the number (Y)on the outside that shift the graph up and down.

Horizontal Shift: (x+3)^-4, (x-3)^2-9

-The horizontal shift is the number in the parenthesis (X) that shift the graph left or right.

Vertical Stretch: 2(x+4)^2-7

-The vertical stretch is what makes the graph narrower, and it is found in the d value at the front of the equation.

Horizontal Stretch: 1/2(x-4)^2+9

-The horizontal stretch is what makes the graph wider or fatter, and is found in the d value at the front of the equation.

Reflections over the x-axis: -(x-3)^2+3

-The reflections over the x-axis is when you have a negative in the d value

The answers to completing the square, the square root method, finding the solution, and quadratic formula can get four different answers:

Roots: Can be Real or Imaginary numbers, this means that the answer can have a negative number in the square root and end with an i.

Ex:x=-3+-i√5-Imaginary

x=+-2√5-Real

Zeros:Can only have Real numbers, with zeros you can actually go and look at it on the calculator, so it can not have a negative in the square root or end with an i. If it does then it will be no solution.

Ex: x=5+-√-503 =No Solution

/2

X-Intercepts:Can be Real numbers only

Ex:x=-5+-√49 =(1/3), -2

/6

Solutions:Can be Real or Imaginary numbers

Ex. x=4+-2√5-Real

x=-3+-i√11-Imaginary

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