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The Quadratic
Formula
Objective: To Solve Quadratic Equations using the Quadratic
Formula, to write a polynomial function in standard form
given its zeros
The Quadratic Formula
• The Quadratic Formula can be used to find the zeros of a
quadratic function or the roots (solutions) of a quadratic
equation
• Remember that a quadratic equation has the form ax2 + bx + c = 0
• The Quadratic Formula is:
b  b2  4ac
x
2a
The Quadratic Formula
• Solve:
2x2 – 3x + 1 = 0
• As this is already in standard form, we can see that:
• a=2
b = -3
c=0
b  b2  4ac
x
2a
(3)  (3)2  4(2)(1)
x
2(2)
x
3 98
4
x
3 1
4
31
4
31
31
x
or x 
4
4
4
2
x  or x 
The solution set is
4
4
{½ , 1}
1
x  1 or x 
2
x
The Quadratic Formula
• Solve:
3x2 + 4x – 2 = 0
• As this is already in standard form, we can see that:
• a=3
b=4
c = -2
b  b2  4ac
x
2a
(4)  (4)2  4(3)(2)
x
2(3)
x
4  16  24
6
4  40
x
6
4  4  10
6
4  2 10
x
6
2  10
x
3
x
The solution set is
 2  10 


3


Finding the Equation of a
Quadratic Function
• Find the zeros of the function:
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x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x – 4 = 0 or
x=4
f(x) = x2 – 3x – 4
x+1=0
x = -1
• If we know the zeros of a function, we can work backwards to
determine the equation of the function.
• Step 1:
• Step 2:
• Step 3:
Use the zeros to determine the factors
If given a value of “a”, write it in front of the factors
Multiply to get the function into standard form
**NOTE: THIS METHOD WORKS ONLY WHEN THE ZEROS
ARE INTEGERS**
Finding the Equation of a
Quadratic Function
• Find the function with zeros -2 and 5
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Given the zeros, we know that x = -2 and x = 5
Working backward, we can find the factors (x + 2) and (x – 5)
The function (in factored form) is f(x) = (x + 2)(x – 5)
There is no given value of “a,” so the function in standard form is
f(x) = x2 – 3x – 10
• Find the function with zeros 1 and 6 and a = 2
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x = 1 and x = 6
Factors (x – 1) and (x – 6)
f(x) = 2(x – 1)(x – 6)
f(x) = 2(x2 – 7x + 6)
f(x) = 2x2 – 14x + 12
Finding the Equation of a
Quadratic Function
• Find the function with zero 4 and a = -3
• Because there is only one zero, and we know that a quadratic
function must have two zeros, 4 must be a double zero
• x=4
and
x=4
• Factors: (x – 4) and (x – 4)
• f(x) = -3(x – 4)2
• f(x) = -3(x2 – 8x + 16)
• f(x) = -3x + 24x – 48
Finding the Equation of a
Quadratic Function
• Find the function with zeros -2 and 6 and y-intercept -24
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x = -2 and x = 6
Factors: (x + 2) and (x – 6)
f(x) = a(x + 2)(x – 6)
To find the value of a, we can substitute the y-intercept
coordinates (0, -24)
-24 = a(0 + 2)(0 – 6)
-24 = a(2)(-6)
-24 = -12a
a=2
f(x) = 2(x + 2)(x – 6)
f(x) = 2(x2 – 4x – 12)
f(x) = 2x2 – 8x – 24
Finding the Equation of a
Quadratic Function
• Find the function with zeros 3/2 and -1 and a = -4
• As the zeros ARE NOT integers, we have to look at the “a”
value at the end of the problem
•
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x = 3/2
and
x = -1
2x = 3
and
x = -1
Factors: (2x – 3) and (x + 1)
f(x) = (2x – 3)(x + 1)
f(x) = 2x2 – x – 3
The “a” value once the factors have been multiplied is 2, when it
should be -4. We need to adjust the function to correct the “a”
value. We do this by multiplying by -2.
• f(x) = -2(2x2 – x – 3)
• f(x) = -4x2 + 2x + 6
Finding the Equation of a
Quadratic Function
• Find the function with zeros -1 and -2 and one point at
(-3, -4)
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x = -1 and x = -2
Factors: (x + 1) and (x + 2)
f(x) = a(x + 1)(x + 2)
To find the value of a, we can substitute the point coordinates
(-3, -4)
-4 = a(-3 + 1)(-3 + 2)
-4 = a(-2)(-1)
-4 = 2a
a = -2
f(x) = -2(x + 1)(x + 2)
f(x) = -2(x2 + 3x + 2)
f(x) = -2x2 – 6x – 4