Algebra 2 Notes Chapter 4 Notes Part 2 1 Concept Byte: Writing Equations From Roots Recall: A SOLUTION is a value of x that makes the quadratic equation ax 2 + bx + c = 0 true. Sometimes this is also called a ZERO of a function or a ROOT of an equation. Synonyms SOLUTION = ZERO = ROOT On a graph A “REAL” SOLUTION = ZERO = ROOT X‐Intercept 2 x 2 + 11x + 5 = 0
(2 x + 1)( x + 5) = 0
x + 5 x − 14 = 0
( x + 7)( x − 2) = 0
Recall: “SOLVE BY FACTORING” or 2 x + 1 = 0
x+7 =0 x−2 =0
2 x = −1
x = −7
x=2
1
x=−
2
2
x +5 = 0 x = −5
We can work backwards from the roots (the solutions) to find the original equation. Example: Write a quadratic equation with each pair of Steps: values as roots. 1. If “c” is a root, then (x –c ) is a factor. 1. ‐7, 2 2. Write your roots as the product of two factors = 0. roots: b, c equation: ( x – b) (x – c) = 0 3. Multiply using FOIL and simplify. 4. Multiply through by a constant to eliminate 2. ‐5, ‐ ½ fractions if needed. Algebra 2 Notes Chapter 4 Notes Part 2 2 3. 2, 10 4. ‐6, 5 5. ‐5, 5 2
, ‐9 3
6. 2 2
, 5 5
8. − ,
7. 3 1
4 2
Algebra 2 Notes Chapter 4 Notes Part 2 3 Putting it all together 4.4‐4.7 4.4 Factoring Patterns: Greatest Common Factor Trinomial a = 1 16 x 2 − 20 x x 2 + 29 x + 100 Difference of Perfect Squares x 2 − 64 25 x 2 − 4 Trinomial a ≠ 1 Perfect Square Trinomial 3 x 2 + 17 x + 10 25 x 2 + 30 x + 9 20 x 2 − 80 “Factor Completely” − x 2 + 3 x − 28 4.5 Solve by Factoring Steps: 1. Get ax 2 + bx + c = 0 2. Factor 3. Set each factor = 0 4. Solve Solve by Factoring x 2 − 144 = 0 x 2 + 6 x − 40 = 0 9 x 2 − 18x = 0 2 x2 + 9 x + 4 = 0 Algebra 2 Notes Chapter 4 Notes Part 2 4 4.6 Solve by Completing the Square Solve by Completing the Square Steps: x 2 − 4 x + 5 = 8 1. Isolate x 2 + bx ⎛b⎞
⎝2⎠
2
2. Add ⎜ ⎟ to both sides 3. Factor and simplify 4. Solve using square roots x 2 + 2 x − 35 = 0 4.7 Solve by Quadratic Formula Solve by Quadratic Formula Steps: If ax 2 + bx + c = 0 2 x2 − x − 4 = 2 8 x 2 + 1 = 7 x −b ± b 2 − 4ac
x=
2a
Then Discriminant: Steps: Find b 2 − 4ac Positive Æ 2 real solutions Zero Æ 1 real solutions Negative Æ 0 real solutions 10 x 2 − x + 9 = 0 Find the Discriminant. State the number of real solutions. 3 x 2 − 6a + 3 = 0 −9 x 2 + 6 x + 11 = 0