The Quadratic Formula.
The solutions of the equation \(~~ax^2+bx+c=0,~~(a \not= 0)~~\) are
\begin{equation*}
\blert{x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}
\end{equation*}
Section 4.1 Quadratic Formula
Subsection 4.1.1 A New Formula
Instead of completing the square every time we solve a new quadratic equation, we can complete the square on the general quadratic equation,
\begin{equation*}
ax^2+bx+c=0,~~~~ (a \not= 0)
\end{equation*}
and obtain a formula for the solutions of any quadratic equation. Here is the resulting formula.
Note 4.1.1.
The formula gives us the solutions of a particular quadratic equation in terms of its coefficients, \(a,~b,\) and \(c\text{.}\) We know that there should be two solutions, and the symbol \(~\blert{\pm}~\) is used to represent the two expressions
\begin{equation*}
x=\dfrac{-b +\sqrt{b^2-4ac}}{2a}~~~\text{and}~~~x=\dfrac{-b - \sqrt{b^2-4ac}}{2a}
\end{equation*}
in a single formula. \(~\alert{\text{[TK]}}~~\)
Checkpoint 4.1.2. QuickCheck 1.
Which of the following is a correct statement of the quadratic formula?
- \(\displaystyle x=-b \pm \dfrac{\sqrt{b^2-4ac}}{2a}\)
- \(\displaystyle x=\dfrac{-b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}\)
- \(\displaystyle x=-b \pm \dfrac{b-\sqrt{4ac}}{2a}\)
- \(\displaystyle x=-b \pm \dfrac{\sqrt{b^2-ac}}{a}\)
To solve a quadratic equation using the quadratic formula, all we have to do is substitute the coefficients \(a,~b,\) and \(c\) into the formula.
Example 4.1.3.
Solve \(~~2x^2 + 1 = 4x\text{.}\)
Solution.
Write the equation in standard form as
\begin{equation*}
2x^2 - 4x + 1 = 0
\end{equation*}
We substitute \(\alert{2}\) for \(a\text{,}\) \(\alert{-4}\) for \(b\text{,}\) and \(\alert{1}\) for \(c\) into the quadratic formula, then simplify.
\begin{align*}
x \amp =\frac{-(\alert{-4})\pm \sqrt{(\alert{-4})^2 - 4(\alert{2})(\alert{1})}}{2(\alert{2})}\\
\amp = \frac{4 \pm \sqrt{8}}{4}
\end{align*}
Remeber that we cannot cancel the 4’s in this expression! \(~\alert{\text{[TK]}}~~\) Using a calculator, we find that the solutions are approximately \(1.7\) and \(0.3\text{.}\) These values are the \(x\)-intercepts of the graph of \(~y = 2x^2 - 4x + 1,~\) as shown in the figure.
Checkpoint 4.1.4. Practice 1.
Use the quadratic formula to solve \(~x^2 - 3x = 1\text{.}\) \(~\alert{\text{[TK]}}~~\)
Hint: Write the equation in standard form.
Substitute \(a = 1\text{,}\) \(b = -3\text{,}\) \(c = -1\) into the quadratic formula.
Simplify.
Solution.
\(x=\dfrac{3\pm \sqrt{13}}{2} \)
Subsection 4.1.2 Applications
We have now seen four different algebraic methods for solving quadratic equations:
- Factoring
- Extraction of roots
- Completing the square
- Quadratic formula
Factoring and extraction of roots are relatively fast and simple, but they do not work on all quadratic equations. The quadratic formula will work on any quadratic equation.
Checkpoint 4.1.5. QuickCheck 2.
Match each equation with the most efficient method of solution.
- \(\displaystyle 6(x-4)^2=120 \)
- \(\displaystyle x^2-9x+20=0\)
- \(\displaystyle 1.4x^2-6.2x+2.5=0\)
- \(\displaystyle x^2-20x=44\)
- Factoring
- Extraction of roots
- Completing the square
- Quadratic formula
Example 4.1.6.
The owners of a day-care center plan to enclose a divided play area against the back wall of their building, as shown below. They have \(300\) feet of picket fence and would like the total area of the playground to be \(6000\) square feet. Can they enclose the playground with the fence they have, and if so, what should the dimensions of the playground be?
Solution.
Suppose the width of the play area is \(x\) feet. Because there are three sections of fence along the width of the play area, that leaves \(300 - 3x\) feet of fence for its length.
The area of the play area should be \(6000\) square feet, so we have the equation
\begin{align*}
\text{width} \times \text{length} \amp = \text{Area}\\
x(300 - 3x) \amp = 6000
\end{align*}
This is a quadratic equation. We use the distributive law and write the equation in standard form.
\begin{align*}
3x^2 - 300x + 6000 \amp= 0\amp\amp \blert{\text{Divide each term by 3.}}\\
x^2 - 100x + 2000 \amp= 0
\end{align*}
The left side of this equation cannot be factored, so we use the quadratic formula with \(a = 1\text{,}\) \(b = -100\text{,}\) and \(c = 2000\text{.}\)
\begin{align*}
x \amp=\frac{-(-100) \pm\sqrt{(-100)^2 - 4(1)(2000)}}{2(1)}\\
\amp= \frac{100 \pm\sqrt{2000}}{2}\approx \frac{100 \pm 44.7}{2}
\end{align*}
When we evalute this last expression, we get two different positive values for the width of the play area, \(x = 72.4\) or \(x=27.6\text{.}\) Both values give solutions to the problem. To find the length of the playground in each case, we substitute \(x\) into \(300-3x.\)
- If the width of the play area is \(72.4\) feet, the length is \(300 - 3(72.4)\text{,}\) or \(82.8\) feet.
- If the width is \(27.6\) feet, the length is \(300 - 3(27.6)\text{,}\) or \(217.2\) feet.
The dimensions of the play area can be 72.4 feet by 82.8 feet, or it can be 27.6 feet by 217.2 feet.
Checkpoint 4.1.7. Practice 2.
The height of a baseball is given by the equation
\begin{gather*}
h=-16t^2+64t+4
\end{gather*}
where \(t\) is the time in seconds. Find two times when the ball is at a height of 20 feet. Round your answers to two decimal places.
Hint: Step 1: Set \(h=20\text{,}\) then write the equation in standard form.
Step 2: Divide each term by \(-16\text{.}\)
Step 3: Use the quadratic formula to solve.
Solution.
0.27 sec, 3.73 sec
Subsection 4.1.3 Complex Numbers
Not all quadratic equations have solutions that are real numbers. For example, when we try to solve the equation \(~x^2+4=0,~\text{,}\) we find
\begin{align*}
x^2 \amp =-4\\
x \amp = \pm \sqrt{-4}
\end{align*}
Although square roots of negative numbers such as \(\sqrt{-4}\) are not real numbers, they occur frequently in mathematics and its applications. Mathematicians in the sixteenth century gave them the name imaginary numbers, which reflected the mistrust with which they were viewed at the time. Today, however, such numbers are well understood and are used routinely by scientists and engineers.
To help us work with imaginary numbers, we define a new number, \(i\), the imaginary unit, whose square is \(-1\text{.}\)
Definition 4.1.8. Imaginary Unit.
\begin{equation*}
\blert{i = \sqrt{-1}}~~~~\text{or}~~~~\blert{i^2=-1}
\end{equation*}
With this new number we define the principal square root of any negative real number as follows.
Imaginary Numbers.
If \(~a \ge 0\text{,}\)
\begin{equation*}
\blert{\sqrt{-a} = \sqrt{-1}\sqrt{a} = i\sqrt{a}}
\end{equation*}
Thus, the square root of any negative real number can be written as the product of a real number and \(i\text{.}\) Every negative real number has two imaginary square roots. For example, the square roots of \(-9\) are \(3i\) and \(-3i\text{.}\) You can verify that
\begin{equation*}
(3i)^2=9i^2=9(-1)=-9~~~\text{and}~~~(-3i)^2=(-3)^2i^2=9(-1)=-9
\end{equation*}
Example 4.1.9.
- \begin{align*} \sqrt{-4} \amp = \sqrt{-1}\sqrt{4}\\ \amp = i \sqrt{4} = 2i \end{align*}
- \begin{align*} \sqrt{-3} \amp = \sqrt{-1}\sqrt{3}\\ \amp = i \sqrt{3} \end{align*}
Checkpoint 4.1.10. Practice 3.
Simplify.
- \(\displaystyle -\sqrt{-36}\)
- \(\displaystyle (5i)^2\)
Solution.
- \(\displaystyle -6i \)
- \(\displaystyle -25 \)
The solutions of many quadratic equations involve imaginary numbers.
Example 4.1.11.
Solve \(~~2x^2-x+2 = 0\text{.}\)
Solution.
For this equation, \(a=\alert{2}\text{,}\) \(b=\alert{-1}\text{,}\) and \(c=\alert{2}\text{.}\) We substitute these values into the quadratic formula to obtain
\begin{gather*}
x=\frac{ -(\alert{-1}) \pm \sqrt{ (\alert{-1})^2 - 4(\alert{2}) (\alert{2})} } { 2(\alert{2}) } = \frac{1 \pm \sqrt{-15} } {4}
\end{gather*}
We write the solutions as
\begin{gather*}
\frac{1 \pm i\sqrt{15}}{4} \qquad \text{or} \qquad \frac{1}{4} \pm \frac{\sqrt{15}}{4}i
\end{gather*}
Because the solutions are not real numbers, the graph of
\begin{gather*}
y=2x^2-x+2
\end{gather*}
has no \(x\)-intercepts, as shown below.
The sum of a real number and an imaginary number is called a complex number.
Checkpoint 4.1.12. Practice 4.
Use extraction of roots to solve \((2x+1)^2+9=0 \text{.}\) Write your answers as complex numbers.
Solution.
\(\dfrac{-1}{2} + \dfrac{3}{2}i\text{,}\) \(\dfrac{-1}{2} - \dfrac{3}{2}i\)
Subsection 4.1.4 Number of \(x\)-Intercepts
The graph of the quadratic equation
\begin{equation*}
y = ax^2 + bx + c
\end{equation*}
may have two, one, or no \(x\)-intercepts, according to the number of distinct real-valued solutions of the equation \(ax^2 + bx + c = 0\text{.}\) For example, consider the three graphs shown at right.
- The graph of\begin{equation*} y = x^2 - 4x + 3 \end{equation*}has two \(x\)-intercepts, because the equation\begin{equation*} x^2 - 4x + 3 = 0 \end{equation*}has two real-valued solutions, \(x = 1\) and \(x = 3\text{.}\)
- The graph of\begin{equation*} y = x^2 - 4x + 4 \end{equation*}has only one \(x\)-intercept, because the equation\begin{equation*} x^2 - 4x + 4 = 0 \end{equation*}has only one (repeated) real-valued solution, \(x = 2\text{.}\)
- The graph of\begin{equation*} y = x^2 - 4x + 6 \end{equation*}has no \(x\)-intercepts, because the equation\begin{equation*} x^2 - 4x + 6 = 0 \end{equation*}has no real-valued solutions.
A closer look at the quadratic formula reveals useful information about the solutions of quadratic equations. The sign of the number under the radical determines how many solutions the equation has. For the three equations above, we calculate as follows:
\begin{alignat*}{3}
\amp y=x^2-4x+3 \amp \amp y=x^2-4x+4 \amp\amp y=x^2-4+6\\
\amp~~\blert{ \text{two } x\text{-intercepts}} \amp \amp ~~\blert{~\text{one } x\text{-intercept}} \amp\amp ~\blert{~\text{no } x\text{-intercepts}} \\
\small{x} \amp \small{=\frac{4\pm\sqrt{(-4)^2-4(1)(3)}}{2}}\quad\amp
\small{x} \amp \small{=\frac{4\pm\sqrt{(-4)^2-4(1)(4)}}{2}}\quad\amp
\small{x}\amp \small{=\frac{4\pm\sqrt{(-4)^2-4(1)(6)}}{2}}\\
\amp =\frac{4\pm\sqrt{\alert{4}}}{2}\amp
\amp =\frac{4\pm\sqrt{\alert{0}}}{2}\amp
\amp =\frac{4\pm\sqrt{\alert{-12}}}{2}\\
\amp~~\blert{ \text{two real solutions}} \amp \amp ~\blert{~\text{one repeated solution}} \amp\amp ~~\blert{~\text{no real solutions}}
\end{alignat*}
Note 4.1.13.
From these examples, we see that the solutions of a quadratic equation always occur in conjugate pairs,
\begin{equation*}
\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}~~~~~~\text{and}~~~~~~\dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}
\end{equation*}
For example, if we know that one solution of a particular quadratic equation is \(3+\sqrt{2}\text{,}\) the other solution must be \(3-\sqrt{2}\text{.}\) If one solution is \(5-3i\text{,}\) the other solution must be \(5+3i\text{.}\)
The expression \(~b^2-4ac~\text{,}\) which appears under the radical in the quadratic formula, is called the discriminant, \(D\text{,}\) of the equation. The value of the discriminant determines the nature of the solutions of the equation. In particular, if the discriminant is negative, the equation has no real-valued solutions; the solutions are complex numbers.
The Discriminant.
The discriminant of a quadratic equation is
\begin{equation*}
\blert{D=b^2-4ac}
\end{equation*}
- If \(D \gt 0\text{,}\) there are two unequal real solutions.
- If \(D = 0\text{,}\) there is one solution of multiplicity two.
- If \(D \lt 0\text{,}\) there are two complex conjugate solutions.
Example 4.1.14.
Use the discriminant to discover how many \(x\)-intercepts the graph has.
- \(\displaystyle y = x^2 - x - 3 \)
- \(\displaystyle y = 2x^2 + x + 1\)
- \(\displaystyle y = x^2 - 6x + 9\)
Solution.
We set \(y=0\) and compute the discriminant for the resulting equation.
- \(D = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 13 \gt 0\text{.}\) The equation has two real, unequal solutions, and the graph has two \(x\)-intercepts.
-
\(D = b^2 - 4ac = 1^2 - 4(2)(1) = -7 \lt 0\text{.}\)The equation has no real solutions, so the graph has no \(x\)-intercepts.
-
\(D = b^2 - 4ac = (-6)^2 - 4(1)(9) = 0\text{.}\)The equation has one real solution of multiplicity two, and the graph has a single \(x\)-intercept.
Checkpoint 4.1.15. QuickCheck 3.
True or False.
- The discriminant is part of the quadratic formula.
- We use the discriminant to calculate the solutions of a quadratic equation.
- If the discriminant is negative, both \(x\)-intercepts of the graph are negative.
- If a quadratic equation won’t factor, its graph has no \(x\)-intercepts.
Checkpoint 4.1.16. Practice 5.
Use the discriminant to discover how many \(x\)-intercepts the graph of each equation has.
- \(\displaystyle y = x^2 + 5x + 7\)
- \(\displaystyle y = -\dfrac{1}{2}x^2 + 4x - 8\)
Solution.
- None: the discriminant is negative.
- One: the discriminant is 0.
Subsection 4.1.5 Solving Formulas
Sometimes it is useful to solve a quadratic equation for one variable in terms of the others.
Example 4.1.17.
Solve \(~~x^2 - xy + y = 2~~\) for \(x\) in terms of \(y\text{.}\)
Solution.
We first write the equation in standard form as a quadratic equation in the variable \(x\text{.}\)
\begin{equation*}
x^2 - yx + (y - 2) = 0
\end{equation*}
Expressions in \(y\) are treated as constants with respect to \(x\text{,}\) so that \(a = \alert{1}\text{,}\) \(b = \alert{-y}\text{,}\) and \(c = \alert{y - 2}\text{.}\) We substitute these expressions into the quadratic formula.
\begin{align*}
x \amp=\frac{-(\alert{-y}) \pm\sqrt{(\alert{-y})^2 - 4(\alert{1})(\alert{y - 2})}}{2(\alert{1})}\\
\amp = \frac{y \pm\sqrt{y^2 - 4y + 8}}{2}
\end{align*}
Checkpoint 4.1.18. Practice 6.
Solve \(2x^2 + kx + k^2 = 1\) for \(x\) in terms of \(k\text{.}\)
Solution.
\(x=\dfrac{-k\pm\sqrt{8-7k^2}}{4} \)
Exercises 4.1.6 Problem Set 4.1
Warm Up
Exercise Group.
For Problems 1 and 2, simplify according to the order of operations.
1.
- \(\displaystyle 8-2\sqrt{36-16}\)
- \(\displaystyle \dfrac{4+4\sqrt{4(16)}}{4}\)
2.
- \(\displaystyle \dfrac{6+3\sqrt{36-4(3)}} {2(3)}\)
- \(\displaystyle \dfrac{\sqrt{10^2+4(11)}-\sqrt{9+3(6)}} {12-\sqrt{1+5(16)}}\)
3.
- Write the formulas for the area and perimeter of a rectangle.
- Find the area and perimeter of each rectangle:
- 4 ft by 6 ft
- 3 ft by 8 ft
4.
- Do all rectangles with the same area have the same perimeter?
- Write expressions for the area and perimeter of each rectangle:
- \(w~\) in by \(~w-3~\) in
- \(w~\) cm by \(~160-2w~\) cm
Skills Practice
Exercise Group.
For Problems 5-10, use the quadratic formula to solve. Round your answers to three decimal places.
5.
\(0=x^2+x-1\)
6.
\(3z^2=4z+1\)
7.
\(0=x^2-\dfrac{5}{3}x+\dfrac{1}{3}\)
8.
\(-5.2z^2+176z+1218=0\)
9.
\(2x^2=7.5x-6.3\)
10.
\(x=\dfrac{1}{2}x^2-\dfrac{3}{4}\)
11.
- Graph \(~y=6x^2-7x-3~\) in the window\begin{gather*} \text{Xmin}=-2~~~~\text{Ymin}=-6\\ \text{Xmax}=3~~~~~\text{Ymax}=4 \end{gather*}
- Estimate the \(x\)-intercepts of the graph.
- Use the quadratic formula to solve the equation\begin{equation*} 6x^2-7x-3=0 \end{equation*}How do the solutions compare to your estimates in part (b)?
12.
- Solve by the quadratic formula\begin{equation*} 2x^2-1=5x \end{equation*}Find decimal approximations for your answers. Round to hundredths.
-
Graph \(~y=2x^2-5x-1~\) in the window\begin{gather*} \text{Xmin}=-1~~~~\text{Ymin}=-5\\ \text{Xmax}=5~~~~~\text{Ymax}=5 \end{gather*}Sketch your graph on the grid.
- How are your solutions from part (a) related to the graph?
13.
- Graph the three equations\begin{align*} y \amp = x^2-6x+5\\ y \amp = x^2-6x+9\\ y \amp = x^2-6x+12 \end{align*}in the window\begin{gather*} \text{Xmin}=-2~~~~\text{Ymin}=-5\\ \text{Xmax}=7.4~~~~~\text{Ymax}=15 \end{gather*}Use the Trace to locat the \(x\)-intercepts of each graph.
- Use the quadratic formula to find the solutions of each equation.\begin{align*} x^2-6x+5 \amp = 0\\ x^2-6x+9 \amp = 0\\ x^2-6x+12 \amp = 0 \end{align*}How are your answers related to the graphs in part (a)?
14.
Use the discriminant to determine the nature of the solutions of each equation.
- \(\displaystyle 3x^2+26=17x\)
- \(\displaystyle 4x^2+23x=19\)
- \(\displaystyle 16x^2-712x+7921=0\)
- \(\displaystyle 0.03x^2=0.05x-0.12\)
15.
Here is one solution of a quadratic equation. Find the other solution, then write a quadratic equation in standard form that has those solutions.
- \(\displaystyle 2+\sqrt{5}\)
- \(\displaystyle 4-3i\)
16.
- What is the sum of the two solutions of the quadratic equation \(~ax^2 + bx + c = 0~\text{?}\) (Hint: The two solutions are given by the quadratic formula.)
- What is the product of the two solutions of the quadratic equation \(~ax^2 + bx + c = 0~\text{?}\) (Hint: Do not try to multiply the two solutions given by the quadratic formula! Think about the factored form of the equation.)
Exercise Group.
For Problems 17-22, use the quadratic formula to solve the equation for the indicated variable.
17.
\(h=4t-16t^2,~~\) for \(t\)
18.
\(A=2w^2+4lw,~~\) for \(w\)
19.
\(s=vt-\dfrac{1}{2}at^2,~~\) for \(t\)
20.
\(3x^2+xy+y^2=2,~~\) for \(y\)
21.
\(S=\dfrac{n^2+n}{2},~~\) for \(n\)
22.
\(A=\pi r^2 + \pi rs,~~\) for \(r\)
Applications
23.
A car traveling at \(s\) miles per hour on a wet road surface requires approximately \(d\) feet to stop, where \(d\) is given by the equation
\begin{equation*}
d = \dfrac{s^2}{12}+\dfrac{s}{2}
\end{equation*}
- Make a table showing the stopping distance, \(d\text{,}\) for speeds of 10, 10, \(\ldots\text{,}\) 100 miles per hour. (Use the Table feature of your calculator.)
- Graph the equation for \(d\) in terms of \(s\text{.}\) Use your table values to help you choose appropriate window settings.
- Insurance investigators at the scene of an accident find skid marks 100 feet long leading up to the point of impact. Write and solve an equation to discover how fast the car was traveling when it put on the brakes. Verify your answer on your graph.
24.
A high diver jumps from the 10-meter springboard. His height in meters above the water \(t\) seconds after leaving the board is given by
\begin{equation*}
h=-4.9t^2+8t+10
\end{equation*}
- Make a table of values showing the diver’s altitude at 0.25-second intervals after he jumps from the springboard. (Use the Table feature of your calculator.)
- Graph the equation. Use your table of values to choose appropriate window settings.
- How long is it before the diver passes the board on the way down?
- How long is it before the diver hits the water?
- Find points on your graph that correspond to your answers to parts (c) and (d).
25.
When you look down from a height, say a tall building or a mountain peak, your line of sight is tangent to the Earth at the horizon, as shown in the figure.
- The radius of the earth is 6370 kilometers. How far can you see from an airplane at an altitude of 10,000 meters? (You will need to use the Pythagorean theorem.)
- How high would the airplane have to be in order for you to see a distance of 300 kilometers?
26.
Refer to the figure in Problem 25.
- Suppose you are standing on top of the Petronas Tower in Kuala Lumpur, 1483 feet high. How far can you see on a clear day? (The radius of the Earth is 3960 miles. Don’t forget to convert the height of the Petronas Tower to miles.)
- How tall a building should you stand on in order to see 100 miles?
27.
The volume of a large aquarium at the zoo is 2160 cubic feet. The tank is 10 feet wide, and its length is 6 feet less than twice its height.
- Sketch the aquarium, and label its dimensions in terms of a variable.
- Write an equation for the the variable.
- Find the dimensions of the aquarium.
28.
You have 72 feet of rope to enclose a rectangular display area against one wall of an exhibit hall. The area enclosed depends upon the dimensions of the rectangle you make. Because the wall makes one side of the rectangle, the length of the rope accounts for only three sides, as shown below.
- Let \(h\) stand for the height of a rectangle, and write an algebraic expression for the base of the rectangle.
- Write an expression for the area of the rectangle.
- If you would like to enclose 640 square feet of display space, what should the dimensions of the rectangle be? (There are two possible solutions.)
- What should the dimensions be if you would like to enclose exactly 600 square feet of space?
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