# PREVIEWSUMMARY OF QUADRATIC EQUATIONS FUNCTIONS QUADRATIC means second

- Slides: 47

PREVIEW/SUMMARY OF QUADRATIC EQUATIONS & FUNCTIONS QUADRATIC – means second power Recall LINEAR – means first power

METHOD 1 - FACTORING l l l 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. 5 x 2 = 15 x 5 x 2 – 15 x = 0 5 x (x – 3) = 0 5 x = 0 OR x – 3 = 0 x = 0 OR x = 3 {0, 3}

METHOD 1 - FACTORING Binomials – Look for Difference of Squares 2. x 2 = 9 x 2 – 9 = 0 Conjugates (x + 3) (x – 3) = 0 x + 3 = 0 OR x – 3 = 0 x = – 3 OR x = 3 {– 3, 3}

METHOD 1 - FACTORING Trinomials – Look for PST (Perfect Square Trinomial) 3. x 2 – 8 x = – 16 x 2 – 8 x + 16 = 0 (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. 2 x 3 – 15 x = 7 x 2 2 x 3 – 7 x 2 – 15 x = 0 (x) (2 x 2 – 7 x – 15) = 0 {-3/2, 0, 5} (x) (2 x + 3)(x – 5) = 0 x = 0 OR 2 x + 3 = 0 OR x – 5 = 0 x = 0 OR x = – 3/2 OR x = 5

METHOD 2 – SQUARE ROOTS OF BOTH SIDES l l l Reorder terms IF needed Works whenever form is (glob)2 = c Take square roots of both sides (Remember you will need a l l sign!) Simplify the square root if needed Solve for x. (Isolate it. )

METHOD 2 – SQUARE ROOTS OF BOTH SIDES 1. x 2 = 9 x= 3 {-3, 3} Note means both +3 and -3! x = -3 OR x = 3

METHOD 2 – SQUARE ROOTS OF BOTH SIDES 2. x 2 = 18

METHOD 2 – SQUARE ROOTS OF BOTH SIDES 3. x 2 = – 9 Cannot 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 {-1, 5} This means: x = 2 + 3 and x = 2 – 3 x = 5 and x = – 1

METHOD 2 – SQUARE ROOTS OF BOTH SIDES Rewrite as (glob)2 = c first if necessary. 5. x 2 – 10 x + 25 = 9 (x – 5)2 = 9 x = 8 and x = 2 {2, 8}

METHOD 2 – SQUARE ROOTS OF BOTH SIDES Rewrite as (glob)2 = c first if necessary. 6. x 2 – 10 x + 25 = 48 (x – 5)2 = 48

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 2 2 Example: 3 x – 6 = x + 12 x 2 x 2 – 12 x – 6 = 0 Simplify and write in standard form: ax 2 + bx + c = 0 x 2 – 6 x – 3 = 0 Set a = 1 by division Note: in some problems a will already be equal to 1.

METHOD 3 – COMPLETE THE SQUARE x 2 – 6 x – 3 = 0 x 2 – 6 x =3 Move constant to other side x 2 – 6 x + 9 = 3 + 9 Add (b/2)2 to both sides Leave space to replace it! This completes a PST! (x – 3)2 = 12 Rewrite as (glob)2 = c

METHOD 3 – COMPLETE THE SQUARE (x – 3)2 = 12 Take square roots of both sides – don’t forget Simplify Solve for x

METHOD 4 – QUADRATIC FORMULA • • 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:

METHOD 4 – QUADRATIC FORMULA Use to solve: 3 x 2 – 6 = x 2 + 12 x Standard Form: 2 x 2 – 12 x – 6 = 0

METHOD 4 – QUADRATIC FORMULA

REVIEW – QUADRATIC FUNCTIONS 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-intercepts 1. Set y or "f(x)" to zero on one side of the equation 2. 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 – QUADRATIC FUNCTIONS Example Problem: f(x) = x 2 – 2 x – 8 a. Opens UP since a = 1 (that is, positive) b. x-intercepts: 0 = x 2 – 2 x – 8 0 = (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

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES 1. x 2 = 121 x = 11 {-11, 11} Note means both +11 and -11! x = -11 OR x = 11

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES 2. x 2 = – 81 Cannot take a square root of a negative. There are NO real number solutions!

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES x 2 = 26 Rewrite as (glob)2 = c first if necessary. 3. 6 x 2 = 156

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES 4. (a – 7)2 = 3

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES Rewrite as (glob)2 = c first if necessary. 5. 9(x 2 – 14 x + 49) = 4 (x – 7)2 = 4/9 {6⅓, 7⅔}

PRACTICE METHOD 2 – SQUARE ROOTS OF BOTH SIDES 6.

PRACTICE METHOD 3 – COMPLETE THE SQUARE 2 Example: 2 b = 16 b + 6 2 b 2 – 16 b – 6 = 0 Simplify and write in standard form: ax 2 + bx + c = 0 b 2 – 8 b – 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 2 – 8 b – 3 = 0 b 2 – 8 b =3 b 2 – 8 b + 16 = 3 +16 Move constant to other side Leave space to replace it! Add (b/2)2 to both sides This completes a PST! (b – 4)2 = 19 Rewrite as (glob)2 = c

PRACTICE METHOD 3 – COMPLETE THE SQUARE (b – 4)2 = 19 Take square roots of both sides – don’t forget Simplify Solve for the variable

PRACTICE METHOD 3 – COMPLETE THE SQUARE Example: 3 n 2 + 19 n + 1 = n - 2 3 n 2 + 18 n + 3 = 0 Simplify and write in standard form: ax 2 + bx + c = 0 n 2 + 6 n + 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 2 + 6 n + 1 = 0 n 2 + 6 n = -1 Move constant to other side Leave space to replace it! n 2 + 6 n + 9 = -1 + 9 Add (b/2)2 to both sides This completes a PST! (n + 3)2 =8 Rewrite as (glob)2 = c

PRACTICE METHOD 3 – COMPLETE THE SQUARE (n + 3)2 = 8 Take square roots of both sides – don’t forget Simplify Solve for the variable

PRACTICE METHOD 3 – COMPLETE THE SQUARE What number “completes each square”? 1. x 2 – 10 x = -3 1. x 2 – 10 x + 25 = -3 + 25 2. x 2 + 14 x =1 2. x 2 + 14 x + 49 = 1 + 49 3. x 2 – 1 x =5 3. x 2 – 1 x + ¼ = 5 + ¼ 4. 2 x 2 – 40 x =4 4. x 2 – 20 x + 100 = 2 + 100

PRACTICE METHOD 3 – COMPLETE THE SQUARE Now rewrite as (glob)2 = c 1. x 2 – 10 x + 25 = -3 + 25 1. (x – 5)2 = 22 2. x 2 + 14 x + 49 = 1 + 49 2. (x + 7)2 = 50 3. x 2 – 1 x + ¼ = 5 + ¼ 3. (x – ½ )2 = 5 ¼ 4. x 2 – 20 x + 100 = 2 + 100 4. (x – 10)2 = 102

PRACTICE METHOD 3 – COMPLETE THE SQUARE Show all steps to solve. 2 ⅓k = 4 k k 2 = 12 k 2 k -⅔ -2 - 12 k =-2 k 2 - 12 k + 36 = - 2 + 36 (k - 6)2 = 34

PRACTICE METHOD 4 – QUADRATIC FORMULA Show all steps to solve & simplify. 2 2 x =x +6 2 2 x – 6 =0

PRACTICE METHOD 4 – QUADRATIC FORMULA Show all steps to solve & simplify. 2 x +x +5=0

PRACTICE METHOD 4 – QUADRATIC FORMULA Show all steps to solve & simplify. x 2 +2 x - 4 = 0

THE DISCRIMINANT – MAKING PREDICTIONS b 2 – 4 ac is called the discriminant Four cases: 1. b 2 – 4 ac positive non-square two irrational roots 2. b 2 – 4 ac positive square two rational roots 3. b 2 – 4 ac zero one rational double root 4. b 2 – 4 ac negative no real roots

THE DISCRIMINANT – MAKING PREDICTIONS Use the discriminant to predict how many “roots” each equation will have. 1. x 2 – 7 x – 2 = 0 49– 4(1)(-2)=57 2 irrational roots 2. 0 = 2 x 2– 3 x + 1 9– 4(2)(1)=1 2 rational roots 3. 0 = 5 x 2 – 2 x + 3 4– 4(5)(3)=-56 no real roots 4. x 2 – 10 x + 25=0 100– 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 = 2 x 2 – x - 6 1– 4(2)(-6)=49 2 rational zeros opens up/vertex below x-axis/2 x-intercepts 2. f(x) = 2 x 2 – x + 6 1– 4(2)(6)=-47 no real zeros 3. y = -2 x 2– 9 x + 6 81– 4(-2)(6)=129 2 irrational zeros 4. f(x) = x 2 – 6 x + 9 opens up/vertex above x-axis/No x-intercepts opens down/vertex above x-axis/2 x-intercepts 36– 4(1)(9)=0 one rational zero opens 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: glob 2 = constant. Examples: 3(x + 2)2=12 or x 2 – 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: x 2 – 8 x = 0 has a GCF 4 x 2 – 12 x + 9 = 0 is a PST x 2 – 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: x 2 – 6 x + 1 = 0

QUADRATIC FORMULA – Derive it by Completing the Square! Start with Standard Form: ax 2 + bx + c = 0

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