Quadratic Equations
Definition of a Quadratic Equation
A quadratic equation in x is an equation that
can be written in the standard form
ax2 + bx + c = 0
where a, b, and c are real numbers with a not
equal to 0. A quadratic equation in x is also
called a second-degree polynomial
equation in x.
The Zero-Product Principle
If the product of two algebraic expressions is
zero, then at least one of the factors is equal
to zero.
If AB = 0, then A = 0 or B = 0.
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Solving a Quadratic Equation by
Factoring
1. If necessary, rewrite the equation in the form
ax2 + bx + c = 0, moving all terms to one side,
thereby obtaining zero on the other side.
2. Factor.
3. Apply the zero−product principle, setting each
factor equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.
Text Example
• Solve 2x2 + 7x = 4 by factoring and then using the
zero−product principle.
Step 1 Move all terms to one side and obtain
zero on the other side. Subtract 4 from both sides
and write the equation in standard form.
2x2 + 7x − 4 = 4 − 4
2x2 + 7x − 4 = 0
Step 2 Factor.
2x2 + 7x − 4 = 0
(2x − 1)(x + 4) = 0
Solution cont.
• Solve 2x2 + 7x = 4 by factoring and then
using the zero−product principle.
Steps 3 and 4 Set each factor equal to
zero and solve each resulting equation.
2x−1=0
or x + 4 = 0
2x=1
x = −4
x = 1/2
Steps 5 check your solution
2
Example
(2x + -3)(2x + 1) = 5
4x2 - 4x - 3 = 5
4x2 - 4x - 8 = 0
4(x2-x-2)=0
4(x - 2)*(x + 1) = 0
x - 2 = 0, and x + 1 = 0
So x = 2, or -1
The Square Root Method
If u is an algebraic expression and d is a
positive real number, then u2 = d has exactly
two solutions.
If u2 = d, then u = √d or u = -√d
Equivalently,
If u2 = d then u = ±√d
Completing the Square
If x2 + bx is a binomial, then by adding (b/2)
2, which is the square of half the coefficient
of x, a perfect square trinomial will result.
That is,
2
x + bx + (b/2)2 = (x + b/2)2
3
Text Example
What term should be added to the binomial x2
+ 8x so that it becomes a perfect square
trinomial? Then write and factor the
trinomial.
The term that should be added is the square of
half the coefficient of x. The coefficient of x
is 8. Thus, (8/2)2 = 42. A perfect square
trinomial is the result.
x2 + 8x + 42 = x2 + 8x + 16 = (x + 4)2
Quadratic Equation
ax 2 + bx + c = 0
Quadratic Formula
− b ± b 2 − 4ac
x=
2a
4
x2 − 8x + 5 = 0
x=
− ( −8) ± (−8) 2 − 4(1)(5)
2(1)
8 ± 64 − 20
2
8 ± 44
x=
2
x=
8 ± 2 11
2
2(4 ± 11)
x=
2
x = 4 ± 11
x=
The Discriminant and the Kinds of Solutions
to ax2 + bx +c = 0
Discriminant
b2 – 4ac
Kinds of solutions
to ax2 + bx + c = 0
b2
Two unequal real solutions
– 4ac > 0
Graph of
y = ax2 + bx + c
Two x-intercepts
b2 – 4ac = 0
One real solution
(a repeated solution)
b2 – 4ac < 0
No real solution;
two complex imaginary
solutions
One x-intercept
No x-intercepts
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The Pythagorean Theorem
The sum of the squares of the lengths of the
legs of a right triangle equals the square of
the length of the hypotenuse.
If the legs have lengths a and b, and the
hypotenuse has length c, then
a2 + b2 = c2
Quadratic Equations
Quadratic Functions
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Graphs of Quadratic Functions
The graph of any quadratic function is called a parabola.
Parabolas are shaped like cups, as shown in the graph below. If
the coefficient of x2 is positive, the parabola opens upward;
otherwise, the parabola opens downward. The vertex (or turning
point) is the minimum or maximum point.
The Standard Form of a
Quadratic Function
• The quadratic function
• f (x) = a(x − h)2 + k,
a≠0
• is in standard form. The graph of f is a parabola
whose vertex is the point (h, k). The parabola is
symmetric to the line x = h. If a > 0, the parabola
opens upward; if a < 0, the parabola opens
downward.
Graphing Parabolas With
Equations in Standard Form
•
1.
2.
3.
4.
5.
To graph f (x) = a(x − h)2 + k:
Determine whether the parabola opens upward or
downward. If a > 0, it opens upward. If a < 0, it opens
downward.
Determine the vertex of the parabola. The vertex is (h,
k).
Find any x-intercepts by replacing f (x) with 0. Solve
the resulting quadratic equation for x.
Find the y-intercept by replacing x with zero.
Plot the intercepts and vertex. Connect these points with
a smooth curve that is shaped like a cup.
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Text Example
• Graph the quadratic function f (x) = −2(x − 3)2 + 8.
Solution We can graph this function by following the steps in the preceding
box. We begin by identifying values for a, h, and k.
Standard form
f (x) = a(x − h)2 + k
a = -2
h=3
k=8
Given equation f (x) = −2(x − 3)2 + 8
Step 1 Determine how the parabola opens. Note that a, the coefficient of
x 2, is -2. Thus, a < 0; this negative value tells us that the parabola opens
downward.
Text Example cont.
Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h =
3 and k = 8, the parabola has its vertex at (3, 8).
Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) = −2(x − 3)2 + 8.
0 = −2(x − 3)2 + 8
(x − 3)2 = 4
(x − 3) = ±2
x − 3 = −2
x=1
Find x-intercepts, setting f (x) equal to zero.
Solve for x. Add 2(x − 3)2 to both sides of
the equation.
2(x − 3)2 = 8
or
or
Divide both sides by 2.
Apply the square root method.
x−3=2
x=5
Express as two separate equations.
Add 3 to both sides in each equation.
The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).
Text Example cont.
Step 4
Find the y-intercept. Replace x with 0 in f (x) = −2(x − 3)2 + 8.
f (0) = −2(0 − 3)2 + 8 = −2(−3)2 + 8 = −2(9) + 8 = −10
The y-intercept is –10. The parabola passes through (0, −10).
Step 5 Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5,
and a y-intercept at –10, the axis of symmetry is the vertical line whose
equation is x = 3.
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The Vertex of a Parabola Whose
Equation Is f (x) = ax 2 + bx + c
• Consider the parabola defined by the
quadratic function
• f (x) = ax 2 + bx + c. The parabola's vertex is
at
 −b  −b  
 , f  
 2a  2a  
Example
Graph the quadratic function f (x) = −x2 + 6x −2.
Solution:
Step 1 Determine how the parabola opens. Note that a, the
coefficient of x 2, is -1. Thus, a < 0; this negative value tells us that the
parabola opens downward.
Step 2 Find the vertex. We know the x-coordinate of the vertex is x =
-b/(2a). We identify a, b, and c to substitute the values into the
equation for the x-coordinate:
x = -b/(2a) = -6/2(-1)=3.
The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of
the function to find the y-coordinate:
y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, the parabola has its vertex at (3,7).
Example
Graph the quadratic function f (x) = −x2 + 6x −2.
• Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) = −x2 + 6x
− 2.
a = −1,b = 6,c = −2
• 0 = −x2 + 6x − 2
x=
=
−b ± b 2 − 4ac
2a
−6 ± 6 2 − 4(−1)(−2)
2(−1)
−6 ± 36 − 8
−2
−6 ± 28 −6 ± 2 7
=
=
−2
−2
= 3± 7
=
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Example
Graph the quadratic function f (x) = −x2 + 6x −2.
• Step 4 Find the y-intercept. Replace x with 0 in f (x) = −x2 + 6x −
2.
•
f (0) = −02 + 6 • 0 − 2 = −2
The y-intercept is –2. The parabola passes through
(0, −2).
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8
Step 5 Graph the parabola.
6
4
2
-10 -8 -6 -4 -2
2
4
6
8 10
-2
-4
-6
-8
-10
Minimum and Maximum:
Quadratic Functions
•
Consider f(x) = ax2 + bx +c.
1. If a > 0, then f has a minimum that occurs at
x = -b/(2a). This minimum value is f(-b/(2a)).
2. If a < 0, the f has a maximum that occurs at
x = -b/(2a). This maximum value is f(-b/(2a)).
Quadratic Functions
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