Because we solve certain equations the same way all the time, .... Examples:
Solve the following using the quadratic formula: .... Solving Word Problems.
Solving Quadratic Equations By Square Root Method
Solving Quadratic Equations By Completing The Square
Consider the equation x2 = a2, which we now solve:
1. Put the equation in a form such that the quadratic and linear terms are on one side of the equation and the constant term is on the other side.
x 2 = a2 x 2 – a2 = 0 (x – a)(x + a) = 0 x–a=0 x+a=0 x=a x = –a x = ±a
2. Make sure that the quadratic coefficient is one. Divide both sides by the coefficient if necessary. 3. Make the critical calculations:
Because we solve certain equations the same way all the time, we can now take a shortcut.
a) b) c) d)
Square Root Rule: You may take the square root of both sides of an equation provided you use ± on one side.
4. Factor one side (which must be a perfect square trinomial) and simplify the other.
This allows us to solve many different types of equations. Ex 1:
x2 = 5 x = ± √5
(x + 2)2 = 5 x + 2 = ± √5 x = –2 ± √5
Ex 2:
5. Take the square roots of both sides of the equation. (Don't forget that one side needs ±) 6. Solve the resulting equation and simplify.
WARNING: If you use the square root rule, you had better remember the ±. Forgetting the ± will not be tolerated!!
Examples: Solve the following equations by the square root method:
a)
x2 = 4
b)
x = ± √4
y2 = 5
c)
y = ± √5
x = ±2
Take the linear coefficient. Take half of it (or divide it by 2). Square it. Add this result to both sides of the equation.
z2 = 18
d)
w2 = –9
z = ± √18
w = ± √–9
z = ±3 √2
w = ±3i
7. Check your answers.
Examples: Solve the following equations by completing the square: a)
x2 – 8x = 25
b)
3x2 + 6x – 15 = 0
x2 – 8x + 16 = 25 + 16
3x2 + 6x = 15
(x – 4)2 = 41
x2 + 2x = 5
x – 4 = ± √41
x2 + 2x + 1 = 5 + 1
x=4± √41
(x + 1)2 = 6 x + 1 = ± √6
e)
i)
x2
= –5
f)
x2
+4=0
g) (x –
3)2
= 25 h) (y +
2)2
x = ± √–5
x2
= –4
x – 3 = ±5
x = ±i √5
x = ± √–4
x=3±5
y = –2 ±3
x = ±2i
x = 8, –2
y = 1, –5
(z – 4)2 = 5 z – 4 = ± √5 z=4± √5
j)
(w + 1)2 = –4 w + 1 = ±2i w = –1 ± 2i
k)
x = –1 ± √6
=9
y + 2 = ±3 c)
2x2 + 2x – 7 = 0
d)
2x2 + 5x – 3 = 0
2x2 + 2x = 7
2x2 + 5x = 3
7 x2 + x = 2
5 3 x2 + 2 x = 2
1 7 1 x2 + x + 4 = 2 + 4
5 25 3 25 x2 + 2 x + 16 = 2 + 16
2x + 1 = ± √3
1 15 (x + 2)2 = 4
5 49 (x + 4)2 = 16
2x = –1 ± √3
1 √15 x+2=± 2
5 7 x + 4 = ±4
1 √ 15 x=–2± 2
5 7 x=–4±4
(2x + 1)2 = 3
x=
–1 ± √ 3 2
1 x = –3, 2
e)
x2 – 24x + 72 = 0
f)
x2 – 3x + 1 = 0
x2 – 24x = –72
x2 – 3x = –1
x2 – 24x + 144 = –72 + 144
9 9 x2 – 3x + 4 = –1 + 4
(x – 12)2 = 72
Examples: Solve the following using the quadratic formula:
a)
3 √ 5 x–2=±2
x = 12 ± 6 √2
x =
3 5 x=2±√ 2
g) 3x2 – 6x + 15 = 0 h) x2 + 2x + 2 = 0 x2 + 2x = –2
x2 – 2x = –13
x22 – 2x = –5
x2 + 2x + 1 = –2+1
x2 – 2x +1 = –13+1
x2 – 2x + 1 = –5+1
(x + 1)2 = –1
(x – 1)2 = –12
(x – 1)2 = –4
x + 1 = ±i
x – 1 = ±2i
x = –1 ± i
x – 1 = ±i √12
x = 1 ± 2i
Briefly stated, the quadratic formula tell us that if ax2 + bx + c = 0 b 2 – 4ac –b ± √ . You should memorize this with a ≠ 0, then x = 2a formula. These are the steps you should use to solve an equation: 1. Put the equation in standard form. This means you must get a zero on one side of the equation. 2. Identify the corresponding values of a, b, and c. b 2 – 4ac –b ± √ with these values 3. Evaluate the expression 2a of a, b, and c. 4. Simplify the expression.
b 2 – 4ac –b ± √ 2a
x =
b 2 – 4ac –b ± √ 2a
–(–5) ± √ (–5) 2 – 4(1)(3) 2(1)
=
–(–1) ± √ (–1) 2 – 4(5)(–4) 2(5)
=
5±√ 25 – 12 2
=
1±√ 1 + 80 10
=
5±√ 13 2
=
1±√ 81 10
x =
5±√ 13 2
=
1 ± 9 10
4 x = 1, – 5
c)
x2 + 2x – 4 = 0 a = 1, b = 2, c = –4
x = = = = =
5. Check your answers. = The quantity b 2 – 4ac is called the discriminant and gives us the following information about the solution of the equation. b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0
a = 5, b = –1, c = –4
x = 1 ± 2i √3
Solving Quadratic Equations By The Quadratic Formula
If If If
5x2 – x – 4 = 0
=
i) x2 – 2x + 13 = 0
3x2 – 6x = –15
b)
a = 1, b = –5, c = 3
3 5 (x – 2)2 = 4
x – 12 = ± √72
x2 – 5x + 3 = 0
then there are two real solutions. then there is one real solution. then there are no real solutions. (there are two complex solutions)
b 2 – 4ac –b ± √ 2a –(2) ±
(2) 2 – 4(1)(–4) √ 2(1)
–2 ± √ 4 + 16 2 –2 ± √ 20 2 –2 ± 2 √ 5 2 2(–1 ± √ 5) 2
d)
3 2 1 1 4x –2x=3 3 1 1 12( 4 x2 – 2 x ) = 12( 3 ) 9x2 – 6x = 4 9x2 – 6x – 4 = 0 a = 9, b = –6, c = –4 x =
b 2 – 4ac –b ± √ 2a
=
–(–6) ± √ (–6) 2 – 4(9)(–4) 2(9)
=
6±√ 36 + 144 18
=
6±√ 180 18
=
6 ± 6√ 5 18
=
6(1 ± √ 5) 18
x = –1 ± √5
x =
1±√ 5 3
Examples: Solve the following equations: e)
1 2 1 1 4x –2x=–4
f)
7x2 – 2x + 1 = 0 a = 7, b = –2, c = 1
1 1 1 4( 4 x2 – 2 x ) = 4(– 4)
b 2 – 4ac –b ± √ x = 2a
x2 – 2x = –1 x2 – 2x + 1 = 0
=
a = 1, b = –2, c = 1
–(–2) ± √ (–2) 2 – 4(7)(1) 2(7)
=
2±√ 4 – 28 14
–(–2) ± √ (–2) 2 – 4(1)(1) = 2(1)
=
2±√ –24 14
2±√ 4 – 4 = 2
=
2 ± 2i √ 6 14
=
2(1 ± i √ 6) 14
x =
=
a)
– 4ac –b ± √ 2a b2
2±√ 0 2
x =
2 = 2
x4 – 7x2 + 12 = 0
b)
(x + 2)2 – (x + 2) – 12 = 0
(x2 – 4)(x2 – 3) = 0
u2 – u – 12 = 0
x2 – 4 = 0
x2 – 3 = 0
(u – 4)(u + 3) = 0
x2 = 4
x2 = 3
u–4=0
u+3=0
x = ±2
x = ± √3
u=4
u = –3
x+2=4
x + 2 = –3
x=2
x = –5
x = 2, –2, √ 3, –√ 3
x = 2, –5
c)
1 ± i√ 6 7
x =1
x2/3 – 4x1/3 – 21 = 0
d)
x – x1/2 – 2 = 0
u2 – 4u – 21 = 0
u2 – u – 2 = 0
(u – 7)(u + 3) = 0
(u – 2)(u + 1) = 0
u–7=0
u+3=0
u–2=0
u+1=0
u=7
u = –3
u=2
u = –1
x1/3
x1/3
x1/2
x1/2 = –1
=7
= –3
x = 73
x = (–3)3
x = 343
x = –27
=2
x = 22 = 4
x = (–1)2 = 1
x=4
x = 343, –27
e) g)
2x2 – 3x + 4 = 0 a = 2, b = –3, c = 4
x =
b 2 – 4ac –b ± √ 2a
–(–3) ± √ (–3) 2 – 4(2)(4) = 2(2)
h)
3x2 + 5 = 0 a = 3, b = 0, c = 5
x =
–b ±
b 2 – 4ac √ 2a
–(0) ± √ (0) 2 – 4(3)(5) = 2(3)
=
3±√ 9 – 32 4
±√ –60 = 6
=
3±√ –23 4
=
x =
3 ± i√ 23 4
±2i √15 6
±i √15 x = 3
x –4 – 7x –2 + 10 = 0
f)
x5 – 16x = 0
u2 – 7u + 10 = 0
x(x4 – 16) = 0
(u – 5)(u – 2) = 0
x(x2 – 4)(x2 + 4) = 0
u–5=0
u –2 = 0
u=5
u=2
x2 = 4
x2 = –4
x –2
x –2
x = ±2
x = ±2i
=5
=2
1 =5 x2
1 =2 x2
1 x2 = 5
1 x2 = 2
1 5 √
1 2 √
x=±
x=±
√5 x = ± 5
√2 x = ± 2
√ 5, ±√ 2 x = ± 5
2
x=0
x2 – 4 = 0 x2 + 4 = 0
x = 0, ±2, ±2i
g)
x4 + 16x2 = –48 x4 + 16x2 + 48 = 0 (x2 + 4)(x2 + 12) = 0 x2 + 4 = 0
x2 + 12 = 0
x2 = –4
x2 = –12
x = ±2i
x = ±i √12 x = ±2i √3
x = ±2i, ±2i √3
Solving Word Problems ( Super Solvers Use C.A.P.E.S.)
Example 2: The sum of the squares of three consecutive positive odd integers is 83. Find the integers.
1. Read the problem carefully. (Reread it several times if necessary) 2.
C ategorize the problem type if possible.
(Is it a problem of numerical expression, distance–rate–time, cost–profit, or simple interest type?)
A
ssign a variable to the 3. Decide what is asked for, and unknown quantity. Label the variable so you know exactly what it represents. 4. Draw a
1st odd integer:
n
2nd odd integer:
n+2
3rd odd integer:
n+4
sum of squares
=
n2 + (n+2)2 + (n+4)2 n2
+
n2
5. Form information provided. 6.
=
83
+ 8n + 16
=
83
3n2 + 12n – 63
=
0
3(n2 + 4n – 21)
=
0
3(n – 3)(n + 7)
=
0
+ 4n + 4 +
Picture, diagram, or chart whenever possible!! an E quation (or inequality) that relates the
n2
Solve the equation (or inequality).
n = 3, –7
7. Check your solution with the wording of the problem to be sure it makes sense. distance–rate–time: cost–profit: simple interest: total cost work problems
distance = rate • time (d = r • t) profit = revenue – cost (P = R – C) interest = principal * rate (i = p • r) total cost = cost / item * # items 1 / A + 1 / B = 1 / together
consecutive integers
n, n+1, n+2, ... or n, n+2, n+4, ...
Pythagorean Theorem
a 2 + b 2 = c2
Example 1: The product of two consecutive integers is 20. Find the integers. 1st integer:
n
2nd
n+1
integer:
83
–7 not positive n
=
3 only solution
integers: 3, 5, & 7
Example 3: The length of a rectangle is 2 meters more than its width. If the diagonal is 10 meters, what are the dimensions of the rectangle?
10
w
product = 20 w+2
Pythagorean theorem n(n + 1)
=
20
n2 + n
=
20
n2 + n – 20
=
0
(n – 4)(n + 5)
=
0
n = 4, –5 n = 4 then n+1 = 5 n = –5 then n + 1 = –4 4 & 5 or –5 & –4
w2 + (w+2)2
=
102
w2 + w2 + 4w + 4
=
100
+ 4w – 96
=
0
2(w2 + 2w – 48)
=
0
2(w + 8)(w – 6)
=
0
2w2
w = –8, 6 –8 make no sense length: 8 meters width: 6 meters
Example 4: In two hours, a motorboat can travel 8 miles down a river and return 4 miles back. If the river flows at a rate of 2 miles per hour, how fast can the boat travel in still water?
d = r
•
t
down 8
x+2
8/(x+2)
up
x–2
4/(x–2)
4
Example 6: A grocer mixes $9.00 worth of Grade A coffee with $12.00 worth of Grade B coffee to obtain 30 pounds of a blend. If Grade A costs 30¢ a pound more than Grade B, how many pounds of each were used?
charge
=
price
•
quantity
Grade A
9
9/x
x
Grade B
12
12/(30 – x)
30 – x
total time = 2 hours 4 8 x + 2+x – 2
=
2
9 x
=
12 3 30 – x + 10
9 10x(30 – x) x
=
3 12 10x(30 – x) 3 0 – x + 10
90(30 – x)
=
120x + 3x(30–x)
2700 – 90x
=
120x + 90x – 3x2
4 8 (x+2)(x–2)x + 2 + x – 2
=
2(x+2)(x–2)
8(x – 2) + 4(x + 2)
=
2(x2 – 4)
8x – 16 + 4x + 8
=
2x2 – 8
3x2 – 300x + 2700
=
0
=
2x2 – 12x
3(x2
– 100x + 900)
=
0
=
2x(x – 6)
3(x – 10)( x – 90)
=
0
=
0, 6
x
=
10, 90
0 x
only x = 10 makes sense 6 mph 10 pounds of Grade A 20 pounds of Grade B
Example 5: Two employees together can prepare a large order in 2 hr. Working alone, one employee takes three hours longer than the other. How long does it take each person working alone? one
x+3 hours
other
x
hours
together
2
hours
Solving Polynomial or Rational Inequalities Never multiply both sides of an inequality by an expression involving the variable. To do so involves a complicated procedure you are not ready to handle yet.
WARNING : Multiplying both sides of an inequality by an expression involving the variable will not be tolerated!! To solve polynomial or rational inequalities:
1 1 x + 3+x
=
1 2
1 1 2x(x + 3)x + 3 + x
=
1 2x(x + 3)2
2x + 2(x + 3)
=
x(x + 3)
=
x2
+ 3x
0
=
x2
–x–6
0
=
(x – 3)(x + 2)
x
=
3, –2
2x + 2x + 6
only x = 3 makes sense 6 hours for one 3 hours for other
Step 1: Put the inequality into Standard Form. (Zero on one side) Step 2: Simplify the other side of the inequality. Step 3: Factor the resulting expression. Step 4: Identify all critical values. (They cause the numerator or denominator to be zero. Step 5: Create a sign chart with values that cause the numerator to be zero marked with “0” and values which cause the denominator to be zero marked with “(*)”. This divides the real line into intervals. Step 6: Check each interval using test points to see if the rational expression is positive or negative. If positive, mark with “+”; if negative, mark with “–”. Step 7: Choose the appropriate interval remembering never to choose any value marked with “(*)”.
Examples: Solve the following inequalities and use Mathematics Writing Style: h) a)
x2 – 2x > 8
2x2 – 13x + 15 ≤ 0
(x – 4)(x + 2) > 0
(2x – 3)(x – 5) ≤ 0
+ 0–– 0 +
+ 0–– 0 +
4
5
3/2
6 x + 2 – 1 ≥0 6 – (x + 2) ≥ 0 x + 2 6 – x – 2 ≥ 0 x + 2
x3 + 4x2 – 5x < 0
4 – x x + 2 ≥ 0
x3 ≥ 4x
d)
x(x2 + 4x – 5) < 0
x3 – 4x ≥ 0
x(x + 5)(x – 1) < 0
x(x2 – 4) ≥ 0 x(x – 2)(x + 2) ≥ 0
– 0 + 0– 0 + -5
0
25 x + 3≥ 7–x 25 x + 3+x–7≥0 (x – 7)(x + 3) 25 ≥0 x + 3 x + 3+
3/2 ≤ x ≤ 5
x < –2 or x > 4
c)
i)
2x2 + 15 ≤ 13x
b)
x2 – 2x – 8 = 0
-2
6 x + 2 ≥ 1
1
– (*) + + 0 – -2 4
25 + x 2 – 4x – 21 ≥0 x + 3 x 2 – 4x + 4 ≥0 x + 3 (x – 2) 2 x + 3 ≥0
– 0 + 0– 0 + -2
x < –5 or 0 < x < 1
0
2
–2 < x ≤ 4
– (*) + + 0 +
–2 ≤ x ≤ 0 or x ≥ 2
-3
2
x > –3
e)
x + 3 x – 4 ≥ 0
x2 – 4 x + 1 ≤0
f)
Graphing Quadratic Functions
(x – 2)(x + 2) ≤0 x + 1
+ 0 – – (*) + 4 -3 x ≤ –3 or x > 4
– 0 + (*) – 0 + -2 -1
2
x ≤ –2 or –1 < x ≤ 2
g)
x + 8 x – 5 + 2 < 0 x + 8 + 2(x – 5) < 0 x – 5 x + 8 + 2x – 10 < 0 x – 5 3x – 2 x – 5 < 0 + 0 – – (*) + 2/3
5
2/3 < x < 5
y = x2
typical graph
y=
x2
+1
typical graph moved up 1 unit
y=
x2
+4
typical graph moved up 4 units
y = x2 – 3
typical graph moved down 3 units
y = x2
typical graph
y = (x – 1)2
typical graph moved right 1 unit
y = (x – 4)2
typical graph moved right 4 units
y = (x + 3)2
y = x2 y=
typical graph moved left 3 units
typical graph
–x2
typical graph inverted
y = –5x2
typical graph much thinner, inverted
y = –0.2x2
typical graph much wider, inverted
Graphing Quadratic Functions A quadratic function is a function whose rule is a quadratic polynomial. That is, it can be written in the form f(x) = ax2 + bx + c,
a ≠ 0.
f(x) = a(x – h)2 + k,
a ≠ 0.
or
The graph of such a quadratic function is a parabola with the following information: if a > 0 up 1) The parabola opens down if a < 0 |a| > 1 thinner if 2) The parabola is w i d e r i f 0 < | a | < 1 y = x2
typical graph
y = 2x2
typical graph slightly thinner
y=
5x2
typical graph much thinner
y = 0.5x2
typical graph slightly wider
y = 0.2x2
typical graph much wider
3) The vertex of the parabola has coordinates (h, k) where b k = f(h) h = – 2a = c – ah2 4) The axis of symmetry of the parabola is the line x = h Remember: If you have the quadratic function given in standard form, you can always rewrite it into the vertex form using the above formulas for h and k!
Note that it is much easier to graph when the function has the form f(x) = a(x – h)2 + k. Here are some suggestions for graphing quadratic functions. Step 1: Write in standard form and identify a, b, and c. Step 2: Calculate h:
b h = – 2a
Step 3: Calculate k:
k = f(h)
OR
k = c – ah2
Step 4: Find all of the intercepts. Step 5: Plot sufficiently many points and make use of the intercepts and the vertex (h, k) Step 6: Draw the smooth graph. Example: Graph f(x) = x2 – 4x – 3. y 8.00
-8.00
8.00
-8.00
Vertex: (2, –7)
Intercepts: (0,–3), (2+ √ 7,0), (2–√ 7,0)
Example: Graph f(x) = –x2 – 2x + 3. Find the vertex and any intercepts.
y 8.00
-8.00
8.00
-8.00
