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Chapter 16
Quadratic Equations
What is a Quadratic Equation?
• A quadratic equation has a term such as x2 and no
higher powers
• Sometimes a term xy can be treated as a quadratic term, but
we will not deal with such problems
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We will learn three methods to solve quadratics
1. The square root property: Take the square root of both sides
2. Completing the square: Make one side a perfect square then
take the square root of both sides
3. The quadratic formula: plug values into a formula
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Section 16.1
Solving by the Square Root Property
To Solve, Take the Square Root of Both Sides
• If x2 = 4, then x = 2 or -2
We write this as x = ± 2
• Note: in general, a quadratic equation has two solutions
The exception is x2 = 0, since 0 = -0
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Examples
• Solve: x2 – 12 = 0
x2 = 12
x =  12 =  2 3
• Solve: 6x2 = 14
x2 =
14
6
x=
=
7
3
7
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More Complicated Equations
• Solve: (2t – 3)2 = 25
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Solution
• (2t – 3)2 = 25
• Take the square root of both sides
2t-3 = ± 5
This means 2t – 3 = 5 OR 2t – 3 = -5
t = 4 or t = -1
Remember there are two solutions for most quadratics!
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Example
Solve: x2 + 5 = 0
Solution
• Solve: x2 + 5 = 0
x2 = -5
Take square root of both sides
x = ± −5
However, the square root of a negative number is not real.
Therefore, our equation has no solutions.
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Examples
• (x + 2)2 = 25
• (3x – 17)2 = 28
• (5x + 38)2 – 9 =0
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Solutions
• (x + 2)2 = 25
x + 2 = ± 5, x = 3, x = -7
• (3x – 17)2 = 28
3x – 17 = ± 2 7
3x = ± 2 7 - 17
x=
−17 ±2 7
3
• (5x + 38)2 – 9 =0
5x + 38 = ± 3
5x = -35 or -41, x = -7 or x =
−41
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Section 16.2
Completing the Square
The Method
• Suppose we have x2 + 6x + 9 = 1
• We can factor the left side: (x + 3) 2 = 1
taking square roots of both sides
x+3=1
x = -4 or x = -2
• Try x2 - 10x + 25 = 9
• Again we can factor (x – 5)2 = 9
x–5=3
x = 8 or x = 3
What if we don’t have a perfect square?
•
For example, we have x2 + 8x = 0
•
We determine what would make a perfect square
•
(x + 4) 2 = x2 + 8x + 16, we need to add 16
What has to be added to get a perfect square?
• x2 – 10x
• x2 + 16x
• 2x2 – 7x
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Solutions
• x2 – 10x
add 25
• x2 + 16x
add 64
• 2x2 – 7x
add
49
4
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Using this Method
To solve 3x2 + 12x + 9 = 0
1. If there is a coefficient multiplying the x2, divide by it
x2+ 4x + 3 = 0
2. Move the number to the other side of the equation
x2+ 4x = -3
3. Determine what has to be added to the left to get a
perfect square. To do so, divide the number multiplying x
and square it; here we have (4/2) = 2, 22 = 4
4. Add that number to both sides
x2+ 4x + 4 = -3 + 4 = 1
5. Form the left side into a perfect square
(x + 2)2 = 1
6. Take the square root of both sides
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Another Worked Example
• x2 + 6x +5 = 0
1. No x2 coefficient
2. Move 5: x2 + 6x = - 5
3. 6 multiplies x, divide by 2 to get 3
4. Add 32 = 9 to both sides: x2 + 6x + 9 = -5 + 9
5. Form the perfect square: (x+3)2 = 4
6. Take the square root of both sides: x + 3 =  2
x = -5, x = -1
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Example with Fractions
•
m2 + 5m/2 = 3/2
1. Done
2. Done
3. 5/2  2 = 5/4, squared = 25/16
4. m2 + 5/4m + 25/16 = 3/2 + 25/1624/16 + 25/16 = 49/16
5. (m + 5/4) 2 = 49/16
6. m + 5/4 = 7/4
m = -2/4 = -1/2 and m = -12/4 = -3
Examples
• 2x2 – 16x = -22
• x2 + 3x – 4 = 0
• 4x2 -20x +7 = 0
Solutions
• 2x2 – 16x = -22
x2 – 8x = - 11, (x-4)2= 16 – 11 = 5, x = 4  5
• x2 + 3x – 4 = 0
x2 + 3x + 3/2 = 4 + 9/4, (x + 3) =  (25/4), x = -3  5/2
• 4x2 -20x +7 = 0
x2 -5x = -7/4 becomes (x – 5/2) 2 = 25/4 – 7/4 = 18/4 = 9/2
x – 5/2 = (9/2) = 3(1/2) = 32 / 2
x = [5 32 ]/2
Examples
• t2 - 2t - 2 = 2
• w2 + 12w + 36 = 1/9
• m2 + 5m/2 = 3/2
• 4v2 – 12v + 7 = 0
• (x – 2)(x + 4) = 7
•
Solutions
• t2 - 2t - 2 = 2
(t – 1) 2 + 1 – 2 = 2; (t – 1) 2 = 3; t – 1 = 3
• w2 + 12w + 36 = 1/9
(w + 6) 2 – 36 + 36 = 1/9; w + 6 =  1/3
• m2 + 5m/2 = 3/2
(m + 5/4) – 25/16 = 3/2
• 4v2 – 12v + 7 = 0
(2v – 3) 2 – 9 + 7 = 0; (2v – 3) = 2
• (x – 2)(x + 4) = 7
x 2 + 2x - 8 = 7; (x – 1) 2 – 8 – 1 = 7; x – 1 = 4
Example
• 5x2 + 30x + 20 = 0
Solution
• 5x2 + 30x + 20 = 0
• x2 + 6x = 4
• (x + 3) 2 = 4 + 9 = 15
• x + 3 = 15
Example
• A rectangular photo is 1.5 times as long as it is tall. If the total
area is 24 cm2, what are its dimensions?
Solution
• A rectangular photo is 1.5 times as long as it is tall. If the total
area is 24 cm2, what are its dimensions?
L = 1.5 W; A = LW
24 = 1.5 W2
24/1.5 = W2
24/1.5 = 16, W = 4
Example
• A circular disk is cut from a square. What is the radius of the
disk if the are of the metal wasted is 124 cm2. the area of a
circle is r2
Solution
• A circular disk is cut from a square. What is the radius of the
disk if the are of the metal wasted is 124 cm2. the area of a
circle is r2
• The length of a side of the square is 2r
• Area of the square is 4r2
• Area of the square less area of the circle is r2 (4 -  ) = 124
• r = sqrt [124/(4 -  ) ]
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Section 16.3
Quadratic Formula
Quadratic Formula History
• Babylonia: clay tablets from 2000 BC
• Used in ancient Babylonia, Egypt, Greece, China, and India
• The Egyptian papyri record it in 2050 BC to 1650 BC
• In India, 8th century BC, there are records of solved quadratic
equations of the form ax2 = c and ax2 + bx = c
Formula
Given ax2 + bx + c = 0
−𝑏 ± 𝑏 2 − 4𝑎𝑐
2𝑎
Derivation
ax2 + bx + c = 0
ax2 + b = -c
x2 + bx/a = -c/a
(x + [b/(2a)] ) 2 = -c/a + [b/(2a)] 2
= - 4ac/(4a2) + b2/(4a2)
= (b2 – 4ac)/(4a2)
x + b/(2a) =  (b2 – 4ac) / 2a
x = [-b  (b2 – 4ac) ] / 2a
Using the Quadratic Formula
• x2 – 5x – 6 = 0
Solution
x2 – 5x – 6 = 0
5  (25 + 24) =
2
5  49 = 5  7
2
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Example
• 4x2 – 4x = 1
Solution
• x = 1 2
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The Quadratic Equation
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The Discriminant
x = -b   (b2 – 4ac)
2a
• The portion under the radical, b2 – 4ac is the discriminant
• It “discriminates” or determines what kind of root we have
– If it is zero, then both roots are the same: (-b  0)/2a =
-2b/a; there is only one root, and it is rational
– If it is > 0, then there are two real roots
• If a perfect square, both roots are rational numbers
• If not, both roots are irrational numbers
– If it is < 0, the both roots are complex numbers
Images
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Two Roots
One Root
0
-20
0
-30
20
Complex
Roots
More on Discriminant
• If the discriminant is less than zero:
– There are no real roots
– The graph does not intersect the x axis
• The discriminant allows us to tell if an equation has a real
solution, “real numbers”, those we can see and touch
Examples; Roots?
• x2 + 4x + 4 = 0
• x2 + 4x + 5 = 0
• x2 + 4x + 3 = 0
Solutions
• Find discriminant: b2 -4ac
• x2 + 4x + 4 = 0; discriminant is 0, one real root
• x2 + 4x + 5 = 0; discriminant is -4, only complex roots
• x2 + 4x + 3 = 0; discriminant is 4, two real roots
Section 16.5
Interval Notation
Notation
46
Examples
Graph and put in interval notation:
•
{x | x ≥ 4}
• {x | x < ̶ 3}
• {x | 1 < x ≤ 4}
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Solutions
•
{x | x ≥ 4}
[4, ∞)
• {x | x < ̶ 3}
(‒∞, ‒3)
• {x | 1 < x ≤ 4}
(1, 4]
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Domain and Range
• In an expression f(x) = y
 The x values are the domain, the things that go into the
equation
 The y values are the range, the things that come out of the
equation
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