Solving Equations with the Quadratic Formula By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula. THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a 0. The solutions of the quadratic equation a x 2 + b x + c = 0 are: Remember that before you apply the quadratic formula to a quadratic equation, you must write the equation in standard form, a x 2 + b x + c = 0. x = – b ± b 2 – 4ac 2a
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Solving Equations with the Quadratic Formula
By completing the square once for the general equation a x 2 + b x + c = 0, you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula.
THE QUADRATIC FORMULA
Let a, b, and c be real numbers such that a 0. The solutions of the quadratic equation a x 2 + b x + c = 0 are:
Remember that before you apply the quadratic formula to a quadratic equation, you must write the equation in standard form, a x 2 + b x + c = 0.
x = – b ± b 2 – 4ac
2a
Solve 2x 2 + x = 5.
SOLUTION
Write original equation. 2x 2 + x = 5
2x 2 + x – 5 = 0 Write in standard form.
Quadratic formula
a = 2, b = 1, c = – 5
Simplify.
Solving a Quadratic Equation with Two Real Solutions
x = –b ± b 2 – 4ac
2a
x = –1 ± 1 2 – 4(2)(–5)
2(2)
x = –1 ± 414
Solve 2x 2 + x = 5.
SOLUTION
The solutions are:
1.35
CHECK Graph y = 2 x 2 + x – 5 and note that the x-intercepts are about 1.35 and about –1.85.
–1.85
Solving a Quadratic Equation with Two Real Solutions
and x =–1 – 41
4–1 + 41
4x =
Solving a Quadratic Equation with One Real Solution
Solve x 2 – x = 5x – 9.
SOLUTION
Write original equation. x 2 – x = 5x – 9
x 2 – 6x + 9 = 0 a = 1, b = –6, c = 9
Quadratic formula
Simplify.
x = 3
The solution is 3.
Simplify.
x = 6 ± 0
2
x = 6 ± (– 6) 2 – 4(1)(9)
2(1)
Solving a Quadratic Equation with One Real Solution
Solve x 2 – x = 5x – 9.
CHECK Graph y = x 2 – 6 x + 9 and note that the only x -intercept is 3. Alternatively, substitute 3 for x in the original equation.
6 = 6
3 2 – 3 = 5(3) – 9?
9 – 3 = 15 – 9?
Solving a Quadratic Equation with Two Imaginary Solutions
Solve –x 2 + 2x = 2.
SOLUTION
Write original equation. –x 2 + 2x = 2
–x 2 + 2x – 2 = 0 a = –1, b = 2, c = –2
Quadratic formula
Simplify.
x = 1 ± i
x = –2 ± 2 i
–2Write using the imaginary unit i.
The solutions are 1 + i and 1 – i.
Simplify.
x = –2 ± – 4
–2
x = –2 ± 2 2 – 4(–1)(–2)
2(–1)
Solving a Quadratic Equation with Two Imaginary Solutions
Solve –x 2 + 2x = 2.
SOLUTION
CHECK Graph y = –x 2 + 2 x – 2 and note that there are no x-intercepts. So, the original equation has no real solutions. To check the imaginary solutions 1 + i and 1 – i, substitute them into the original equation. The check for 1 + i is shown.
2 = 2
–(1 + i ) 2 + 2(1 + i ) = 2?
–2i + 2 + 2i = 2?
In the quadratic formula, the expression b 2 – 4 a c under the radical sign is called the discriminant of the associated equation a x 2 + b x + c = 0.
Consider the quadratic equation a x 2 + b x + c = 0.
NUMBER AND TYPE OF SOLUTIONS OF A QUADRATIC EQUATION
discriminant
You can use the discriminant of a quadratic equation to determine the equation’s number and type of solutions.
If b 2 – 4 a c > 0, then the equation has two real solutions.
Solving Equations with the Quadratic Formula
If b 2 – 4 a c = 0, then the equation has one real solution.
If b 2 – 4 a c < 0, then the equation has two imaginary solutions.
–b ± b 2 – 4ac2a
x =
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
SOLUTION
x 2 – 6 x + 10 = 0
x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i
a x + b x + c = 0
EQUATION DISCRIMINANT SOLUTION(S)
b 2 – 4a c
(–6) 2 – 4(1)(10) = – 4
Using the Discriminant
x = –b ± b 2 – 4a c
2a
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
x 2 – 6 x + 10 = 0
x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i
a x + b x + c = 0
EQUATION DISCRIMINANT SOLUTION(S)
b 2 – 4a c
(–6) 2 – 4(1)(10) = – 4
Using the Discriminant
x 2 – 6 x + 9 = 0 (–6) 2 – 4(1)(9) = 0 One real: 3
x 2 – 6 x + 9 = 0
SOLUTION
x = –b ± b 2 – 4a c
2a
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
x 2 – 6 x + 10 = 0
x 2 – 6 x + 10 = 0 Two imaginary: 3 ± i
a x + b x + c = 0
EQUATION DISCRIMINANT SOLUTION(S)
b 2 – 4a c
(–6) 2 – 4(1)(10) = – 4
Using the Discriminant
x 2 – 6 x + 9 = 0 (–6) 2 – 4(1)(9) = 0 One real: 3
x 2 – 6 x + 9 = 0 x 2 – 6 x + 8 = 0
x 2 – 6 x + 8 = 0 (–6) 2 – 4(1)(8) = 4 Two real: 2, 4
SOLUTION
x = –b ± b 2 – 4a c
2a
In the previous example you may have noticed that the number of real solutions of x 2 – 6 x + c = 0 can be changed just by changing the value of c.
y = x 2 – 6 x + 10
y = x 2 – 6 x + 9
y = x 2 – 6 x + 8
Graph is above x-axis (no x-intercept)
Graph touches x-axis (one x-intercept)
Graph crosses x-axis (two x-intercepts)
Using the Discriminant
A graph can help you see why this occurs. By changing c, you can move the graph of y = x 2 – 6x + c up or down in the coordinate plane.
If the graph is moved too high, it won’t have an x-intercept and the equation x 2 – 6 x + c = 0 won’t have a real-number solution.
Using the Quadratic Formula in Real Life
Previously you studied the model h = –16t 2 + h 0 for the height of an object that is dropped. For an object that is launched or thrown, an extra term v 0 t must be added to the model to account for the object’s initial vertical velocity v 0.
Models
h = –16 t 2 + h 0
h = –16 t 2 + v 0 t + h 0
Object is dropped.
Object is launched or thrown.
Labels
h = height
t = time in motion
h 0 = initial height
v 0 = initial vertical velocity
(feet)
(seconds)
(feet)
(feet per second)
Using the Quadratic Formula in Real Life
The initial velocity of a launched object can be positive, negative, or zero.
v 0 > 0 v 0 < 0 v 0 = 0
If the object is launched upward, its initial vertical velocity is positive (v 0 > 0).
If the object is launched downward, its initial vertical velocity is negative (v 0 < 0).
If the object is launched parallel to the ground, its initial vertical velocity is zero (v 0 = 0).
Solving a Vertical Motion Problem
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. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?
SOLUTION
h = –16 t 2 + v 0 t + h 0 Write height model.
h = 5, v 0 = 45, h 0=6
Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v
0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of
t for which h = 5.
5 = –16 t 2 + 45t + 6
0 = –16 t 2 + 45t + 1 a = –16, b = 45, c = 1
Solving a Vertical Motion Problem
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. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?
t – 0.022 or t 2.8
Quadratic formula
a = –16, b = 45, c = 1
Use a calculator.
SOLUTION
Since the baton is thrown (not dropped), use the model h = –16t 2 + v 0 t + h 0 with v
0 = 45 and h 0 = 6. To determine how long the baton is in the air, find the value of
t for which h = 5.
Reject the solution – 0.022 since the baton’s time in the air cannot be negative. The baton is in the air for about 2.8 seconds.
