Download ffpc solving quad notes prob packet

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FFPC…Solving Quadratic Equations
In general, a quadratic expression looks like this: ax2 + bx + c, where a ≠ 0.
There are 5 ways to solve for x in a quadratic equation: a) factor it and use the zero
product property (zpp), b) symbolically get x by itself, c) use the quadratic formula, d)
complete the square, e) use tables/graphs.
We will focus on the first three ways to solve. Here we go.
Factor/ZPP
Step 1) Make sure the equation equals 0 (move numbers/variables around).
2) Factor the equation (remember to pull out any greatest common
factors first). If the equation does not factor, then this method is
worthless. Move on to another method.
3) Once in factored form, figure out what will make each factor equal
to 0… these numbers are your solutions.
Example: (x+3)(x-8) = 0 (this equation is already equal to 0 and it is
already factored). To make each factor equal to 0, x must be -3 or 8. These two
numbers are my solutions.
Symbolically
Step 1) This method only works if there is an x2 term and NO OTHER x. If
there is another x, this method is worthless. Move on to another method.
2) Isolate the x2 term. Move numbers/variables away from the x2 term
to get it by itself.
3) Square root both sides of the equation. When you take the square root
of x2 it becomes the absolute value of x, which looks like this: |x|.
Remember, when solving an absolute value problem, set up two cases, one
positive and one negative.
Example:
x2 + 5 = 21
x2
= 16 (subtract 5 from both sides; x2 term is now isolated)
|x| = 4 (took the square root of both sides)
x
= 4 or -4 (two answers when solving for absolute value)
Quadratic Formula
Step 1) This method ALWAYS works, but it has many details to keep track of.
2) Make sure the equation equals 0 (move numbers/variables around
first if need be).
3) Plug in numbers into the formula below (remember, a quadratic
equation looks like this: ax2 + bx + c = 0, where a ≠ 0): “a” is the
coefficient of the x2 term, “b” is the coefficient of the x term, “c” is the
constant term. The ± symbol simply means there are two cases, one is the
plus (+) case, and one is the minus (-) case.
Example:
•
2
Use the Quadratic Formula to solve x – 4x – 8 = 0.
Looking at the coefficients, I see that a = 1, b = –4, and c
simplify:
= –8. I'll plug them into the Formula, and
Then the solution is
The above solution has two answers: 2 + 2√3 and 2 – 2√3. You can leave the
answer like it is or you can use the decimal answers of 5.464 and -1.464.
Looks fun, huh??? Now, you try these methods on the following problems…
3) x 2 = 21
4) a 2 = 4
Solve using the Factor/ZPP method
2
m 2 + 7 = 88
5) x + 89)
= 28
6) 2n 2 = 10)
−144−5x = −500
a. x 2 − 36 = 0
b. x 2 − 16x + 64 = 0
c. x(2x − 6) = 0
2
−7n 2 = −448
7) −6m 2 11)
= −414
ANSWERS: 6, -6
2
8) 7x 2 = 12)
−21 −2k = −162
ANSWER: 8
ANSWERS: 0, 3
Solve using the Symbolic method
IV.
the song!).
x 2 −Solve
5 = by
73 using the Quadratic Formula (sing
16n 2 = 49
a.
b.10)
9) m 2 + 713)
= 88
−5x 214)
= −500
11) −7n 2 = −448
ANSWERS: 8.83, -8.83
12) −2k 2 = −162
ANSWERS: 10,
-10
-1-
Solve using the Quadratic Formula method
13) x − 5 = 73
14) 16n 2 = 49
2
-1-
ANSWERS: -3, 2
Kuta Software - Infinite Algebra 1
Use
the notes above
to help you
with the following…
Solving
Quadratic
Equations
with Square
Name_________________________
Roots
Solve each equation by
taking square roots.
symbolically.
1) k 2 = 76
2) k 2 = 16
3) x 2 = 21
4) a 2 = 4
5) x 2 + 8 = 28
6) 2n 2 = −144
7) −6m 2 = −414
8) 7x 2 = −21
9) m 2 + 7 = 88
10) −5x 2 = −500
11) −7n 2 = −448
12) −2k 2 = −162
13) x 2 − 5 = 73
14) 16n 2 = 49
-1-
Date________________
Solving Quadratic Equations by Factoring
Date________________ P
Solve each equation by factoring.
1) (k + 1)(k − 5) = 0
2) (a + 1)(a + 2) = 0
3) (4k + 5)(k + 1) = 0
4) (2m + 3)(4m + 3) = 0
5) x 2 − 11x + 19 = −5
6) n 2 + 7n + 15 = 5
7) n 2 − 10n + 22 = −2
8) n 2 + 3n − 12 = 6
9) 6n 2 − 18n − 18 = 6
10) 7r 2 − 14r = −7
Using the Quadratic Formula
Date________________ Period____
Solve each equation with the quadratic formula.
1) m 2 − 5m − 14 = 0
2) b 2 − 4b + 4 = 0
3) 2m 2 + 2m − 12 = 0
4) 2x 2 − 3x − 5 = 0
5) x 2 + 4x + 3 = 0
6) 2x 2 + 3x − 20 = 0
7) 4b 2 + 8b + 7 = 4
8) 2m 2 − 7m − 13 = −10
-1-