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PREVIEW/SUMMARY OF QUADRATIC EQUATIONS
& FUNCTIONS
QUADRATIC – means second power
Recall LINEAR – means first power
METHOD 1 - FACTORING
Set equal to zero Factor Use the Zero Product Property to solve
(Each factor with a variable in it could be equal to zero.)
METHOD 1 - FACTORING
Any # of terms – Look for GCF factoring first!
1. 5x2 = 15x5x2 – 15x = 0
5x (x – 3) = 0
5x = 0 OR x – 3 = 0
x = 0 OR x = 3
{0, 3}
METHOD 1 - FACTORING
Binomials – Look for Difference of Squares
2. x2 = 9x2 – 9 = 0
(x + 3) (x – 3) = 0
x + 3 = 0 OR x – 3 = 0
x = – 3 OR x = 3
{– 3, 3}
Conjugates
METHOD 1 - FACTORING
Trinomials – Look for PST (Perfect Square Trinomial)
3. x2 – 8x = – 16x2 – 8x + 16 = 0
(x – 4) (x – 4) = 0
x – 4 = 0 OR x – 4 = 0
x = 4 OR x = 4
{4 d.r.}
Double Root
METHOD 1 - FACTORING
Trinomials – Look for Reverse of Foil
4. 2x3 – 15x = 7x2
2x3 – 7x2 – 15x = 0
(x) (2x2 – 7x – 15) = 0
(x) (2x + 3)(x – 5) = 0
x = 0 OR 2x + 3 = 0 OR x – 5 = 0
{-3/2, 0, 5}
x = 0 OR x = – 3/2 OR x = 5
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
Reorder terms IF needed Works whenever form is (glob)2 = c Take square roots of both sides (Remember
you will need a sign!) Simplify the square root if needed Solve for x. (Isolate it.)
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
1. x2 = 9
x = 3
Note means both +3 and -3!
x = -3 OR x = 3
{-3, 3}2x = 9
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
2. x2 = 18
2x = 18
x = 9 2
x = 3 2
or x = 3 2 x = 3 2
OR3 2
-3 2 3 2,
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
3. x2 = – 9
2x = -9
= -9xCannot take a square root of
a negative. There are NO real number solutions!
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
4. (x-2)2 = 9
2x - 2 = 9
x -2= 3x = 2 3
This means: x = 2 + 3 and x = 2 – 3x = 5 and x = – 1
{-1, 5}
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
Rewrite as (glob)2 = c first if necessary.
5. x2 – 10x + 25 = 9
x = 8 and x = 2
{2, 8}(x – 5)2 = 9
2x -5 = 9
x -5= 3x = 5 3
METHOD 2 – SQUARE ROOTS OF BOTH SIDES
Rewrite as (glob)2 = c first if necessary.
6. x2 – 10x + 25 = 48(x – 5)2 = 48
2x -5 = 16 3
x -5= 4 3
x = 5 4 3
OR5 4 3
5 - 4 3,5+ 4 3
METHOD 3 – COMPLETE THE SQUARE
• Goal is to get into the format: (glob)2 = c• Method always works, but is only
recommended when a = 1 or all the coefficients are divisible by a
• We will practice this method repeatedly and then it will keep getting easier!
METHOD 3 – COMPLETE THE SQUARE
Example: 3x2 – 6 = x2 + 12x
2x2 – 12x – 6 = 0Simplify and write in
standard form: ax2 + bx + c = 0
x2 – 6x – 3 = 0 Set a = 1 by division Note: in some problems
a will already be equal to 1.
METHOD 3 – COMPLETE THE SQUARE
(x – 3)2 = 12 Rewrite as (glob)2 = c
x2 – 6x – 3 = 0Move constant to other side Leave space to replace it!
x2 – 6x = 3
x2 – 6x + 9 = 3 + 9 Add (b/2)2 to both sides This completes a PST!
METHOD 3 – COMPLETE THESQUARE
Solve for x
Take square roots of both sides – don’t forget
Simplify
(x – 3)2 = 12
x -3 2 3
x 3 2 3
4 3x -3 2x -3 12
OR3 2 3 3 - 2 3 3+2 3,
METHOD 4 – QUADRATICFORMULA
• This is a formula you will need to memorize!• Works to solve all quadratic equations• Rewrite in standard form in order to identify
the values of a, b and c.• Plug a, b & c into the formula and simplify!• QUADRATIC FORMULA:
2-b± b -4acx =2a
METHOD 4 – QUADRATICFORMULA
Use to solve: 3x2 – 6 = x2 + 12x
2-b± b -4acx =2a
Standard Form: 2x2 – 12x – 6 = 0
2-(-12)± (-12) -4(2)(-6)x =2(2)
METHOD 4 – QUADRATICFORMULA
12± 144+48 12± 192x =4 4
12± 64 3x =4
12±8 3x =4
3 ± 2 3
2-(-12)± (-12) -4(2)(-6)x =2(2)
a. The graph is a parabola. Opens up if a > 0 and down if a < 0.
b. To find x-intercepts: – may have Zero, One or Two x-intercepts1. Set y or "f(x)" to zero on one side of the equation2. Factor & use the Zero Product Prop to find TWO x-intercepts
c. To find y-intercept, set x = 0. Note f(0) will equal c. I.E. (0, c) d. To find the coordinates of the vertex (turning pt):
1. x-coordinate of the vertex comes from this formula: 2. plug that x-value into the function to find the y-coordinate
e. The axis of symmetry is the vertical line through vertex: x =
REVIEW – QUADRATICFUNCTIONS
b2a
b2a
REVIEW – QUADRATICFUNCTIONS
Example Problem: f(x) = x2 – 2x – 8
a. Opens UP since a = 1 (that is, positive)
b. x-intercepts: 0 = x2 – 2x – 80 = (x – 4)(x + 2)(4, 0) and (– 2, 0)
c. y-intercept: f(0) = (0)2 – 2(0) – 8 (0, – 8)
d. vertex:
e. axis of symmetry: x = 1
-2- , 1, f(1) 1, -92(1)
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
1. x2 = 121
x = 11
Note means both +11 and -11!
x = -11 OR x = 11
{-11, 11}2x = 121
2. x2 = – 81
Cannot take a square root of a negative. There are NO
real number solutions!
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
2x = -81
= -81x
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
Rewrite as (glob)2 = c first if necessary.
3. 6x2 = 156
2x 26 =
x2 = 26
x = 26
OR 26 - 26 + 26,
4. (a – 7)2 = 3
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
2a -7 = 3
a-7= 3a =7 3
OR7 ± 3 7 - 3 , 7 + 3
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
Rewrite as (glob)2 = c first if necessary.
5. 9(x2 – 14x + 49) = 4
{6⅓, 7⅔}
(x – 7)2 = 4/9
2 4x -7
9 =
OR
2 23 3
2 23 3
x -7 = x =7
x =7 x =7
PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES
6. 21 94 64
=0t
21 9t =4 64
9
162t
2 21 9=
4 64
94 t16
t
OR
3 3 3
4 4 4± - ,
PRACTICE METHOD 3 – COMPLETE THE SQUARE
Example: 2b2 = 16b + 6
2b2 – 16b – 6 = 0Simplify and write in
standard form: ax2 + bx + c = 0
b2 – 8b – 3 = 0 Set a = 1 by division Note: in some problems
a will already be equal to 1.
PRACTICE METHOD 3 – COMPLETE THE SQUARE
(b – 4)2 = 19 Rewrite as (glob)2 = c
b2 – 8b – 3 = 0Move constant to other side Leave space to replace it!
b2 – 8b = 3
b2 – 8b + 16 = 3 +16 Add (b/2)2 to both sides This completes a PST!
PRACTICE METHOD 3 – COMPLETE THE SQUARE
Solve for the variable
Take square roots of both sides – don’t forget
Simplify
(b – 4)2 = 19
2b - 4 19
19b-4
b 4 19
OR4 19 4 - 19 4+ 19,
PRACTICE METHOD 3 – COMPLETE THE SQUARE
Example: 3n2 + 19n + 1 = n - 2
3n2 + 18n + 3 = 0Simplify and write in
standard form: ax2 + bx + c = 0
n2 + 6n + 1 = 0 Set a = 1 by division Note: in some problems
a will already be equal to 1.
PRACTICE METHOD 3 – COMPLETE THE SQUARE
(n + 3)2 = 8 Rewrite as (glob)2 = c
n2 + 6n + 1 = 0Move constant to other side Leave space to replace it!
n2 + 6n = -1
n2 + 6n + 9 = -1 + 9 Add (b/2)2 to both sides This completes a PST!
PRACTICE METHOD 3 – COMPLETE THE SQUARE
Solve for the variable
Take square roots of both sides – don’t forget
Simplify
(n + 3)2 = 8
2n+3 8
4 2n+3
n -3 2 2 OR-3 2 2 -3 - 2 2 -3+2 2,
PRACTICE METHOD 3 – COMPLETE THE SQUARE
1. x2 – 10x = -3
What number “completes each square”?1. x2 – 10x + 25 = -3 + 25
2. x2 + 14 x = 1 2. x2 + 14 x + 49 = 1 + 49
3. x2 – 1x = 5 3. x2 – 1x + ¼ = 5 + ¼
4. 2x2 – 40x = 4 4. x2 – 20x + 100 = 2 + 100
PRACTICE METHOD 3 – COMPLETE THE SQUARE
1. x2 – 10x + 25 = -3 + 25
Now rewrite as (glob)2 = c1. (x – 5)2 = 22
2. x2 + 14 x + 49 = 1 + 49 2. (x + 7)2 = 50
3. x2 – 1x + ¼ = 5 + ¼ 3. (x – ½ )2 = 5 ¼
4. x2 – 20x + 100 = 2 + 100 4. (x – 10)2 = 102
PRACTICE METHOD 3 – COMPLETE THE SQUARE
2k - 6 34
k - 6 34
k 6 34
⅓k2 = 4k - ⅔Show all steps to solve.
k2 = 12k - 2
k2 - 12k = - 2
k2 - 12k + 36 = - 2 + 36
(k - 6)2 = 34
6 34 6 34,
PRACTICE METHOD 4 – QUADRATIC FORMULA
1 1 + 48
4
1 49
4
2x2 = x + 6Show all steps to solve & simplify.
2x2 – x – 6 =0
1 1 - 4(2)(-6)
2(2)
and
3- , 22
1 7 -6 8
4 4 4
PRACTICE METHOD 4 – QUADRATIC FORMULA
1 1 - 4(1)(5)
(2)(1)
x2 + x + 5 = 0
Show all steps to solve & simplify.
1 1- 20 1 -19
2 2
not a real number
PRACTICE METHOD 4 – QUADRATIC FORMULA
-2 4 - 4(1)(-4)
(2)(1)
-2 4+16 -2 20
2 2
2 -1 5-2 4 5 -2 2 5
2 2 2(1)
x2 +2x - 4 = 0
Show all steps to solve & simplify.
-1 5
THE DISCRIMINANT – MAKING PREDICTIONS
Four cases:b2 – 4ac is called the discriminant
1. b2 – 4ac positive non-square two irrational roots
2. b2 – 4ac positive square two rational roots
3. b2 – 4ac zero one rational double root
4. b2 – 4ac negative no real roots
THE DISCRIMINANT – MAKING PREDICTIONS
Use the discriminant to predict how many “roots” each equation will have.
1. x2 – 7x – 2 = 0
2. 0 = 2x2– 3x + 1
3. 0 = 5x2 – 2x + 3
4. x2 – 10x + 25=0
49–4(1)(-2)=57 2 irrational roots
9–4(2)(1)=1 2 rational roots
4–4(5)(3)=-56 no real roots
100–4(1)(25)=0 1 rational double root
THE DISCRIMINANT – MAKING PREDICTIONS about Parabolas
The “zeros” of a function are the x-intercepts on it’s graph. Use the discriminant to predict how many x-intercepts each parabola will have and where the vertex is located.
1. y = 2x2 – x - 6
2. f(x) = 2x2 – x + 6
3. y = -2x2– 9x + 6
4. f(x) = x2 – 6x + 9
1–4(2)(-6)=49 2 rational zerosopens up/vertex below x-axis/2 x-intercepts
1–4(2)(6)=-47 no real zerosopens up/vertex above x-axis/No x-intercepts
81–4(-2)(6)=129 2 irrational zerosopens down/vertex above x-axis/2 x-intercepts
36–4(1)(9)=0 one rational zeroopens up/vertex ON the x-axis/1 x-intercept
THE DISCRIMINANT – MAKING PREDICTIONS
Note the proper terminology:
The “zeros” of a function are the x-intercepts on it’s graph. Use the discriminant to predict how many x-intercepts each parabola will have.
The “roots” of an equation are the x values that make the expression equal to zero.
Equations have roots. Functions have zeros which are the x-intercepts on it’s graph.
FOUR METHODS – HOW DO I CHOOSE?
Some suggestions:
Quadratic Formula – works for all quadratic equations, but first look for a “quicker” method. You must rewrite into standard form before using Quad Formula! Don’t forget to simplify square roots and use value of the discriminant to predict the number of roots or zeros.
Square Roots of Both Sides – use when the problem can easily be put into the form: glob2 = constant.
Examples: 3(x + 2)2=12 or x2 – 75 = 0
FOUR METHODS – HOW DO I CHOOSE?
Some suggestions:
Factoring – doesn’t always work, but IF you can see the factors, this is probably the quickest method.
Examples: x2 – 8x = 0 has a GCF4x2 – 12x + 9 = 0 is a PSTx2 – x – 6 = 0 is easy to FOIL
Complete the Square – It always works, but if you aren’t quick at arithmetic with fractions, then this method is best used when a = 1 and b is even (so no fractions).
Example: x2 – 6x + 1 = 0
QUADRATIC FORMULA – Derive it by Completing the Square!
2x + xb
a+
c
a= 0
2 bx + x + =
a
c-a
22 2
2 2
b b+ +
b cx +
4a 4ax = -
a a
Start with Standard Form:ax2 + bx + c = 0
2
2
2b c
x = -b 4a•
+2 4aa 4a•a
2
22b - 4ab
x + =2
c
a 4a
2bbx +
- 4
2a
ac= ±
2a2b - 4ac
x = - ±b
2a 2a
2
2
2b b - 4a
±c
x + =2a 4a
2-b ± b - 4acx =
2a