Download Solve Quadratic Equations

Algebra Notes
Solving Quadratic Functions and Equations
Solve Quadratic Equations
Skill: solve quadratic equations by factoring.
A.SSE.A.2 Interpret the structure of expressions. Use the structure of an expression to identify ways
to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares
that can be factored as (x2 – y2)(x2 + y2).
A.SSE.B.3a Write expressions in equivalent forms to solve problems. Choose and produce an
equivalent form of an expression to reveal and explain properties of the quantity represented by
the expression.★ Factor a quadratic expression to reveal the zeros of the function it defines.
A.REI.B.4b Solve equations and inequalities in one variable. Solve quadratic equations by inspection
(e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula and factoring,
as appropriate to the initial form of the equation. Recognize when the quadratic formula gives
complex solutions and write them as a ± bi for real numbers a and b.
Review: Factoring Quadratic Trinomials into Two Binomials (Using the “ac method” or “splitting the
middle term”.) NOTE: You may want to see Factor by Splitting the Middle Term notes AND
Factoring Quadratic Trinomial notes on www.rpdp.net.
Factoring ax 2  bx  c, a  1
Ex 20: Factor x 2  7 x  12 .
Find two integers such that their product is 12 and their sum is 7 .
4 and  3
Write the two binomials as a product.
 x  4 x  3
Factoring ax 2  bx  c, a  1
Ex 21: Factor 2 x 2  7 x  3 .
Step One: Multiply a  c .
23  6
Step Two: Find two integers such that their product is a  c  6 and their sum is b  7 .
6 and 1
Step Three: Rewrite (“split”) the middle term as a sum of two terms using the numbers from Step Two.
2 x 2  6 x  1x  3
Step Four: Factor by grouping. Group the first 2 terms and last 2 terms and factor out the GCF from each
pair.
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Algebra Notes
2x
2
Solving Quadratic Functions and Equations
 6 x   1x  3
 2 x  x  3  1 x  3
Step Five: If Step Four was done correctly, there should be a common binomial factor. Factor this
binomial out and write what remains from each term as the second binomial factor.
 2 x  1 x  3
Ex 22: Factor 5x 2  7 x  2 .
Step One: Multiply a  c .
5  2  10
Step Two: Find two integers such that their product is a  c  10 and their sum is b  7 . 2 and  5
Step Three: Rewrite (“split”) the middle term as a sum of two terms using the numbers from Step Two.
5x 2  2 x  5x  2
Step Four: Factor by grouping. Group the first 2 terms and last 2 terms and factor out the GCF from each
pair.
5x
2
 2 x    5 x  2 
 x  5x  2   15x  2 
Step Five: If Step Four was done correctly, there should be a common binomial factor. Factor this
binomial out and write what remains from each term as the second binomial factor.
5x  2 x  1
Recall: Special Factoring Patterns
Difference of Two Squares:
Perfect Square Trinomial:
SolvingQuadratics
a 2  b2   a  b  a  b 
a 2  2ab  b2   a  b 
2
a 2  2ab  b2   a  b 
2
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Algebra Notes
Solving Quadratic Functions and Equations
Zero Product Property: If the product of two factors is 0, then one or both of the factors must equal 0.
Ex 23: Solve the equation  x  3 x  1  0 using the zero product property.
Since one or both of the factors must equal 0, we will solve the two equations x  3  0 and x  1  0 .
x3 0
x 1  0
x3
x  1
Solutions: x  1, 3
Solving a Quadratic Equation by Factoring
Ex 24: Solve the equation x 2  5x  6 by factoring.
Step One: Write the equation in standard form.
x 2  5x  6  0
Step Two: Factor the quadratic.
 x  3 x  2  0
Step Three: Set each factor equal to zero and solve.
x3 0 x2  0
x3
x2
x  2, 3
Note: Check this answer by graphing on the calculator.
Ex 25: Solve the equation 2 x 2  4 x  8   x 2  x .
3x 2  5 x  8  0
Step One: Write the equation in standard form.
Step Two: Factor the quadratic using the “ac method”.
a  c  24
3x 2  8 x  3x  8  0
b  5
x  3x  8  1  3x  8  0
8 and 3
 3x  8 x  1  0
3x  8  0 x  1  0
8
x
x  1
3
Step Three: Set each factor equal to zero and solve.


8
3
The solutions can be written in set notation:  1, 
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Algebra Notes
Solving Quadratic Functions and Equations
Ex 26: Solve the equation 25  30 y  9 y 2 .
9 y 2  30 y  25  0
Step One: Write the equation in standard form.
Step Two: Factor the quadratic.
3 y 
2
 30 y   5
2
Note: 2  5 3 y   30 y
 3 y  5
2
0
3y  5  0
Step Three: Set each factor equal to zero and solve.
y
5
3
5
3
The solution can be written in set notation:  
Zero(s) of Quadratic Functions: the x-value(s) where the function intersects the x-axis
To find the zero(s), factor the quadratic and set each factor equal to 0.
Note: We can graph quadratic functions by plotting the zeros. The vertex is halfway between the zeros.
Ex 27: Find the zero(s) of the quadratic function y   x 2  2 x  3 and graph the parabola.
Step One: Factor the quadratic polynomial.
Step Two: Set each factor equal to 0 and solve.
Step Three: Find the coordinates of the vertex.
y    x 2  2 x  3
y    x  3 x  1
x  3  0 x 1  0
x  3
x 1
x
3  1 2

 1
2
2
y    1  2  1  3  1  2  3  4
2
Step Four: Plot the points and sketch the parabola.
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Algebra Notes
Solving Quadratic Functions and Equations


You Try: Solve the quadratic equation 5t 2  25  4 t 2  6 by factoring.
QOD: What must be true about a quadratic equation before you can solve it using the zero product
property?
Sample Practice Question(s):
1. What is the solution set for the following equation?
x 2  10 x  9  0
A. {–9, –1}
B. {–9, 1}
C. {–1, 9}
D. {1, 9}
2. Which of the following equations has roots of –7 and 4?
A.
 x  7  x  4  0
B.
 x  7  x  4  0
C.
 x  7  x  4  0
D.
 x  7  x  4  0
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Algebra Notes
Solving Quadratic Functions and Equations
Skill: solve quadratic equations by completing the square.
A.SSE.A.2 Interpret the structure of expressions. Use the structure of an expression to identify ways
to rewrite it.
A.SSE.B.3b Write expressions in equivalent forms to solve problems. Complete the square in a
quadratic expression to reveal the maximum or minimum value of the function it defines.
A.REI.B.4a Solve equations and inequalities in one variable. Solve quadratic equations in one
variable. Use the method of completing the square to transform any quadratic equation in x into
an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula
from this form.
Review: Factoring a Perfect Square Trinomial
a 2  2ab  b2   a  b 
2
a 2  2ab  b2   a  b 
2
NOTE: You may want to review Completing the Square Notes on www.rpdp.net.
Completing the Square: writing an expression of the form x 2  bx as a perfect square trinomial in order
to factor it as a binomial squared
2
b
To complete the square of x  bx , we must add   .
2
2
Teacher Note: Algebra Tiles work well to illustrate completing the square. See Page 729 for an activity.
Ex 28: Find the value of c such that x 2  10 x  c is a perfect square trinomial.
 10 
b  10 , therefore we must add 
  25  c to complete the square.
 2 
2
Note: x 2  10 x  25   x  5
2
Solving a Quadratic Equation by Completing the Square
Ex 29: Solve x 2  8x  7  0 by completing the square.
x 2  8x  7  0
x 2  8 x  7
Step One: Rewrite to make the lead coefficient 1.
Step Two: Take the constant term to the other side.
2
2
b
Step Three: Complete the square (add   to both sides).
2
Step Four: Factor the perfect square trinomial.
SolvingQuadratics
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8
x 2  8 x     7 
2
8
 
2
x 2  8 x  16  9
2
 x  4  9
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2
Algebra Notes
Solving Quadratic Functions and Equations
 x  4
Step Five: Take the square roots of both sides.
 9
x4 3
x  4  3 x  4  3
x  1
x  7
Step Six: Solve for the variable.
The solution set is
2
7, 1 .
Check your answer by factoring.
Ex 30: Solve 2 x 2  12 x  4  0 by completing the square.
2 x 2 12 x 4 0

 
2
2
2 2
2
x  6x  2  0
x2  6x  2
Step One: Rewrite to make the lead coefficient 1.
Step Two: Take the constant term to the other side.
2
Step Three: Complete the square (add   to both sides).
6
x  6x     2 
2
Step Four: Factor the perfect square trinomial.
x 2  6 x  9  11
2
 x  3  11
b
2
2
2
 x  3
Step Five: Take the square roots of both sides.
3 
2
 11
x  3  11
Step Six: Solve for the variable.
The solution set is
2
6
 
2
x  3  11
x  3   11
x  3  11
x  3  11

11, 3  11
Ex 31: Solve x 2  x  3  0 by completing the square.
x2  x  3  0
x2  x  3
Step One: Rewrite to make the lead coefficient 1.
Step Two: Take the constant term to the other side.
 1 
 1 
x  x     3   
 2 
 2 
2
2
2
b
Step Three: Complete the square (add   to both sides).
2
Step Four: Factor the perfect square trinomial.
SolvingQuadratics
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1 13

4 4
2
1  13

x  
2
4

x2  x 
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Algebra Notes
Solving Quadratic Functions and Equations
2
1
13

x  
2
4

Step Five: Take the square roots of both sides.
x
1
13

2
2
1
13

2
2
1  13
x
2
x
Step Six: Solve for the variable.
The solutions are x 
1
13

2
2
1  13
x
2
x
1  13
.
2
You Try: Solve by completing the square.
4 x 2  20 x  11  0
2
b
QOD: Describe why adding   to x 2  bx makes it a perfect square trinomial.
2
Sample CCSD Common Exam Practice Question(s):
What are the roots (solutions) of x 2  x  1  0 ?
A.
A.
1 
1 

3, 1  3

5, 1  5
 1  5 1  5 
,
B. 

2 
 2
1  5 1  5 
,
C. 

2 
 2
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Algebra Notes
Solving Quadratic Functions and Equations
Skill: solve quadratic equations using the quadratic formula.
A.SSEA..2 Interpret the structure of expressions. Use the structure of an expression to identify ways
to rewrite it.
A.REI.B.4a Solve equations and inequalities in one variable. Use the method of completing the
square to transform any quadratic equation in x into an equation of the form (x – p)2 = q that has
the same solutions. Derive the quadratic formula from this form.
A.REI.B.4b Solve equations and inequalities in one variable. Solve quadratic equations by
inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula
and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic
formula gives complex solutions and write them as a ± bi for real numbers a and b.
A.REI.A.1 Understand solving equations as a process of reasoning and explain the reasoning.
Explain each step in solving a simple equation as following from the equality of numbers asserted
at the previous step, starting from the assumption that the original equation has a solution.
Construct a viable argument to justify a solution method.
Notes:
Deriving the Quadratic Formula by Completing the Square
Solve the quadratic equation ax 2  bx  c  0 by completing the square.
ax 2 bx c 0
  
a
a a a
b
c
x2  x   0
a
a
b
c
x2  x  
a
a
Step One: Rewrite so that the lead coefficient is 1.
Step Two: Take the constant term to the other side.
b
2
2
Step Three: Complete the square (add   to both sides).
2
c
 b 
   
2
a
a
 
x2 
b
x
a
x2 
b
b2
4ac  b2
x 2 
a
4a
4a 2
 b 
 
 2a 
b  b2  4ac

x


 
2a 
4a 2

2
Step Four: Factor the perfect square trinomial.
b 
b2  4ac

x




2a 
4a 2

2
Step Five: Take the square roots of both sides.
x
SolvingQuadratics
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b
b2  4ac

2a
4a 2
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2
Algebra Notes
Solving Quadratic Functions and Equations
x
b
b2  4ac

2a
2a
x
b  b2  4ac
2a
Step Six: Solve for the variable.
x
b  b2  4ac

2a
2a
b  b 2  4ac
2a
x
The Quadratic Formula: To solve a quadratic equation in the form ax 2  bx  c  0 , use the formula
x
b  b2  4ac
.
2a
Note: To help memorize the quadratic formula, sing it to the tune of the song “Pop Goes the Weasel”.
Ex 32: Solve the quadratic equation x 2  8 x  1 using the quadratic formula.
Step One: Rewrite in standard form (if necessary).
x 2  8x  1  0
Step Two: Identify a, b, and c.
a  1, b  8, c  1
Step Three: Substitute the values into the quadratic formula.
b  b2  4ac
x
2a
x
x
Step Four: Simplify.
The solution set is
4 
  8 
 8
2 1
2
 4 11
8  64  4 8  60 8  2 15


 4  15
2
2
2

15,4  15
Ex 33: Solve the quadratic equation 5x  1  6 x 2 using the quadratic formula.
Step One: Rewrite in standard form (if necessary).
6 x 2  5x  1  0
Step Two: Identify a, b, and c.
a  6, b  5, c  1
x
Step Three: Substitute the values into the quadratic formula.
x
SolvingQuadratics
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b  b2  4ac
2a
  5 
 5  4  6 1
2  6
2
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Algebra Notes
Solving Quadratic Functions and Equations
5  25  24 5  49 5  7


12
12
12
5  7 12
5  7 2 1
x

 1 x 


12
12
12
12 6
x
Step Four: Simplify.


1
6
The solution set is  1, 
Ex 34: Solve the quadratic equation 12 x  4  9 x 2  0 using the quadratic formula.
Step One: Rewrite in standard form (if necessary).
9 x 2  12 x  4  0
Step Two: Identify a, b, and c.
a  9, b  12, c  4
x
Step Three: Substitute the values into the quadratic formula.
x
x
Step Four: Simplify.
b  b2  4ac
2a
 12  
12   4  9  4 
2 9
2
12  144  144 12  0 12
2



18
18
18
3
 2
 3
The solution set is    .
Ex 35: Solve the quadratic equation 2 x 2  2 x  3  0 using the quadratic formula.
Step One: Rewrite in standard form (if necessary).
2 x2  2 x  3  0
Step Two: Identify a, b, and c.
a  2, b  2, c  3
x
Step Three: Substitute the values into the quadratic formula.
x
b  b2  4ac
2a
  2  
 2   4  2  3
2  2
2
2  4  24 2  20

4
4
There is no real solution to the quadratic equation because 20 is not a real number.
Step Four: Simplify.
SolvingQuadratics
x
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Algebra Notes
Solving Quadratic Functions and Equations
You Try: Solve the equation 6 x  3  2 x 2 using the quadratic formula.
QOD: Write a conjecture about how the radicand in the quadratic formula relates to the number of
solutions that a quadratic equation has.
Sample Practice Question(s):
1. What are the roots (solutions) of x 2  x  1  0 ?
A.
B.
1 
1 

3, 1  3

5, 1  5
 1  5 1  5 
,
C. 

2 
 2
1  5 1  5 
,
D. 

2 
 2
2. Which of the following is the correct use of the quadratic formula to find the solutions
of the equation 2 x 2  7 x  5 ?
7 

A. 


 7   4  2  5 7   7   4  2  5 
,

2  2
2  2

7 

B. 


 7 
2
2

2
 4  2  5  7 
,
2  2
 7 
2
 4  2  5  

2  2



 7 

C. 


 7   4  2  5 7   7   4  2  5 
,

2  2
2  2

 7 

D. 


 7   4  2  5 7   7   4  2  5 
,

2  2
2  2

2
2

SolvingQuadratics
2
2

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Algebra Notes
Solving Quadratic Functions and Equations
Discriminant: the discriminant of the quadratic equation ax 2  bx  c  0 is b2  4ac
Note: The discriminant is the radicand of the quadratic formula!
Determining the Number of Real Solutions of a Quadratic Equation Using the Discriminant
Teacher Note: Students should have come up with this in the QOD.
If b2  4ac  0 , then there are no real solutions.
If b2  4ac  0 , then there is one solution.
If b2  4ac  0 , then there are two real solutions.
Ex 36: Determine the number of real solutions that the equations have.
1.
3x 2  2 x  1
Rewrite the equation in standard form.
Find the discriminant.
Determine the number of real solution(s).
2.
a  3, b  1, c  1
b2  4ac   1  4  31  1  12  11
2
b2  4ac  11  0 , so there are no real solutions .
4  5x  x 2
Rewrite the equation in standard form.
Find the discriminant.
Determine the number of real solution(s).
3.
3x 2  2 x  1  0
x 2  5x  4  0
a  1, b  5, c  4
b2  4ac   5  4 1 4   25  16  41
2
b2  4ac  41  0 , so there are two real solutions .
9 x 2  12 x  4  0
Rewrite the equation in standard form.
Find the discriminant.
Determine the number of real solution(s).
SolvingQuadratics
9 x 2  12 x  4  0
a  9, b  12, c  4
b2  4ac   12  4  9  4   144  144  0
2
b2  4ac  0 , so there is one real solution .
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Algebra Notes
Solving Quadratic Functions and Equations
Determining the Number of x-Intercepts of a Quadratic Function Using the Discriminant
Because the x-intercepts of y  ax 2  bx  c are the same as the zeros of the equation
ax 2  bx  c  0 , we can use the discriminant to determine the number of x-intercepts that a
quadratic function has.
Ex 37: Sketch the graph of a quadratic function with a negative discriminant.
Because the discriminant, b2  4ac  0 , the function will have no x-intercept.
A sample answer is shown in the graph.
Note: Any parabola which does not intersect the x-axis is an acceptable answer.
Application Problem
Ex 38: A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet
above the ground and has an initial vertical velocity of 45 feet per second. This can be modeled
by the equation h  16t 2  45t  6 , where h is the height (in feet) and t is the time (in seconds).
The twirler wants her baton to reach at least 40 feet. Will the baton reach that height?
Substitute h  12 .
40  16t 2  45t  6
Write in standard form.
0  16t 2  45t  34
Find the discriminant.
b2  4ac   45  4  16 34   2025  2176  151
a  16, b  45, c  34
2
Since the discriminant is less than 0, this equation has no real solution. Therefore, the baton could not
reach 40 feet.
How high will the baton reach?
Graph the function h  16t 2  45t  6 .
Find the maximum (vertex).
The baton will reach approximately 37.64 ft .
You Try: Find values for c so that the equation will have no real solution, one real solution, and two real
solutions.
x 2  3x  c  0
QOD: Write a quadratic equation which can be factored. Find its discriminant.
Teacher Note: Have students share their answers to the QOD and allow students to make a conjecture for
how to determine if a quadratic polynomial is factorable using the discriminant. (It must be a perfect
square.)
SolvingQuadratics
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Algebra Notes
Solving Quadratic Functions and Equations
Sample Practice Question(s):
The graph of y   x 2  x  12 has how many x-intercepts?
A. 12
B. 2
C. 1
D. 0
Transformations of quadratic functions
1. How would the graph of the function
a.
b.
c.
d.
be affected if the function were changed to
?
The graph would shift 5 units to the left.
The graph would shift 5 units down.
The graph would shift 5 units up.
The graph would shift 3 units down.
2. How would the graph of the function
a.
b.
c.
d.
be affected if the function were changed to
?
The graph would shift 2 units up.
The graph would shift 5 units up.
The graph would shift 2 units to the right.
The graph would shift 5 unit down.
3. How would you translate the graph of
to produce the graph of
a. translate the graph of
down 4 units
b. translate the graph of
c. translate the graph of
up 4 units
d. translate the graph of
right 4 units
left 4 units
4. Compare the graph of
with the graph of
.
a. The graph of g(x) is wider.
b. The graph of g(x) is narrower.
c. The graph of g(x) is translated 6 units down from the graph of f(x).
d. The graph of g(x) is translated 6 units up from the graph of f(x).
5. Compared to the graph of
, the graph of
a. narrower and translated down
b. narrower and translated up
SolvingQuadratics
is _______.
c. wider and translated down
d. wider and translated up
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Algebra Notes
Solving Quadratic Functions and Equations
6. Four bowls with the same height are constructed using quadratic equations as their shapes. Which
bowl has the narrowest opening?
a.
c. Bowl
Bowl
b.
d. Bowl
Bowl
7. What is the parent function for
a.
b.
?
c.
d.
8. Use the quadratic function f(x) = (x + 3)(x – 1), to graph the function.
y
a.
c.
–10 –8
–6
–4
10
8
8
6
6
4
4
2
2
–2
–2
2
4
6
8
10
–4
–6
–4
–2
–2
–4
–6
–6
–8
–8
–10
–10
d.
y
–6
–10 –8
x
–4
b.
–10 –8
y
10
10
8
8
6
6
4
4
2
2
2
4
6
8
10
x
4
6
8
10
x
2
4
6
8
10
x
y
10
–2
–2
2
–10 –8
–6
–4
–2
–2
–4
–4
–6
–6
–8
–8
–10
–10
9. Use the quadratic formula to solve the equation
Describe how you can use the graph
of the equation
to verify your solutions of the equation
Sketch this
graph and verify your solutions.
SolvingQuadratics
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Algebra Notes
Solving Quadratic Functions and Equations
10. Explain the zero-product property and why it is useful. Give an example to illustrate your
explanation.
Sample answer: The zero-product property states that if a product of factors is equal to zero, then one or
more of the factors is equal to zero. For example, in the equation
the polynomial
can be factored as (x + 7)(x – 2). Since this factored polynomial is equal to zero, then the
value of at least one of the factors is 0. This fact is useful because you can easily solve the equations
and
to find the solutions x = –7 and x = 2.
Construct viable arguments for the following reflection questions.
1. Describe how to use the discriminant to find the number of real solutions to a quadratic
equation.
2. Choose a method to solve x 2  5x  4  0 and explain why you chose that method.
3. Describe how the discriminant can be used to determine if an object will reach a given
height.
4. Explain why there are no solutions to the quadratic equation x 2  9 .
5. Describe how to estimate the solutions of 4  x 2  16 . What are the approximate solutions?
6. How do you find the zeros of a function from its graph?
7. Describe how to find the axis of symmetry of a quadratic function if its graph crosses the x
axis. Describe how to find the axis of symmetry of a quadratic function if its graph does not
cross the x axis?
8. Explain how to graph a quadratic function.
9. What do the vertex and zeros represent in the following situation. As Joe dives into his pool,
his height in feet above the water can be modeled by the function f ( x)  16 x 2  16 x  12
where x is the time in seconds after he begins diving. Find the maximum height of the dive
and the time it takes Joe to reach this height. Then find how long it takes him to reach the
water.
10. Explain two ways to solve x 2  x  6  0 . How are these two methods similar?
11. Describe the relationships among the solutions, the zeros, and the x-intercepts of
y  x 2  4 x  12 .
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