Section 6.2 Solving Quadratic Equations
¶Not every quadratic equation can be solved by factoring or by extraction of roots. For example, the expression \(x^2 + x  1\) cannot be factored, so the equation \(x^2 + x  1 = 0\) cannot be solved by factoring. For other equations, factoring may be difficult. In this section we learn two methods that can be used to solve any quadratic equation.
Subsection Squares of Binomials
In Section 2.1 we used extraction of roots to solve equations of the form
\begin{equation*}
a( px + q)^2 + r = 0
\end{equation*}
where the left side of the equation includes the square of a binomial, or a perfect square. We can write any quadratic equation in this form by completing the square.
Consider the following squares of binomials.
Square of binomial \((x+p)^2\) 
\(p\) 
\(2p\) 
\(p^2\) 
1. \((x+\alert{5})^2=x^2+10x+25\) 
\(\alert{5}\) 
\(2(\alert{5})=10\) 
\(\alert{5}^2=25\) 
2. \((x\alert{{}{}3})^2=x^26x+9\) 
\(\alert{3}\) 
\(2(\alert{3})=6\) 
\(\alert{3}^2=9\) 
3. \((x\alert{{}{}12})^2=x^224x+144\) 
\(\alert{12}\) 
\(2(\alert{12})=24\) 
\(\alert{12}^2=144\) 
In each case, the square of the binomial is a quadratic trinomial,
\begin{equation*}
(x + p)^2= x^2 + 2px + p^2
\end{equation*}
Note that the coefficient of the linear term, \(2p\text{,}\) is twice the constant in the binomial, and the constant term of the trinomial, \(p^2\text{,}\) is its square.
We would like to reverse the process and write a quadratic expression as the square of a binomial. For example, what constant term can we add to
\begin{equation*}
x^2  16x
\end{equation*}
to produce a perfect square trinomial? Compare the expression to the formula above:
\begin{equation*}
\begin{alignedat}{4}
x^2 \amp {}+{} \amp 2px \amp {}+{} \amp p^2 \amp {}={} \amp (x + p)^2 \\
x^2 \amp {}{} \amp 16x \amp {}+{} \amp \text{?} \amp {}={} \amp (x + \text{?})^2 \\
\end{alignedat}
\end{equation*}
We see that
\begin{equation*}
2p = 16,~~\text{ so }~~p = \frac{1}{2}(16) = \alert{8}
\end{equation*}
and
\begin{equation*}
p^2 = (8)^2 = \alert{64}
\end{equation*}
We substitute these values for \(p^2\) and \(p\) into the equation to find
\begin{equation*}
x^2  16x + \alert{64} = (x \alert{8})^2
\end{equation*}
Notice that in the resulting trinomial, the constant term is equal to the square of onehalf the coefficient of \(x\text{.}\) In other words, we can find the constant term by taking onehalf the coefficient of \(x\) and then squaring the result. Adding a constant term obtained in this way is called completing the square.
Example 6.18
Complete the square by adding an appropriate constant; write the result as the square of a binomial.
 \(x^2  12x + {}\underline{\hspace{2.727272727272727em}}\)
 \(x^2 + 5x + {}\underline{\hspace{2.727272727272727em}}\)
Solution
 Onehalf of \(12\) is \(6\text{,}\) so the constant term is \((6)^2\text{,}\) or \(36\text{.}\) We add \(36\) to obtain
\begin{equation*}
x^2  12x \alert{{}+{}36}=(x  6)^2~~~~~~~ \blert{\underset{p^2 = (6)^2 = 36}{\stackrel {p = \frac{1}{2}(12) = 6}{}}}
\end{equation*}
 Onehalf of \(5\) is \(\dfrac{5}{2}\text{,}\) so the constant term is \(\left(\dfrac{5}{2}\right)^2\text{,}\) or \(\dfrac{25}{4}\text{.}\) We add \(\dfrac{25}{4}\) to obtain
\begin{equation*}
x^2 +5x \alert{{}+{}\frac{25}{4}}=\left(x +\frac{5}{2}\right)^2~~~~~~~ \blert{\underset{p^2 = (\frac{5}{2})^2 = \frac{25}{4}}{\stackrel {p = \frac{1}{2}(5) = \frac{5}{2}}{}}}
\end{equation*}
Checkpoint 6.20
Complete the square by adding an appropriate constant; write the result as the square of a binomial.
 \(x^2  18x{} + {}\underline{\hspace{2.727272727272727em}}{}= (x{}\underline{\hspace{2.727272727272727em}} )^2\)
 \(x^2 +9x{} + {}\underline{\hspace{2.727272727272727em}}{}= (x{}\underline{\hspace{2.727272727272727em}} )^2\)
Hint
For part (a): \(p=\frac{1}{2}(18)={}\underline{\hspace{2.727272727272727em}}, p^2={}\underline{\hspace{2.727272727272727em}}\)
For part (b): \(p=\frac{1}{2}(9)={}\underline{\hspace{2.727272727272727em}}, p^2={}\underline{\hspace{2.727272727272727em}}\)
Answer
\(x^2  18x + 81 = (x  9)^2\)
\(x^2+9x+\dfrac{81}{4}=\left(x+\dfrac{9}{2}\right)^2 \)
Subsection Solving Quadratic Equations by Completing the Square
Now we will use completing the square to solve quadratic equations. First, we will solve equations in which the coefficient of the squared term is 1. Consider the equation
\begin{equation*}
x^2  6x  7 = 0
\end{equation*}
and follow the steps to find the solutions.
Step 1 Begin by moving the constant term to the other side of the equation, to get
\begin{equation*}
x^2  6x{}\underline{\hspace{2.727272727272727em}} = 7
\end{equation*}
Step 2 Now complete the square on the left. Because
\begin{equation*}
p = \frac{1}{2}(6) = 3 ~~~\text{ and } ~~~ p^2 = (3)^2 = 9
\end{equation*}
we add \(9\) to both sides of our equation to get
\begin{equation*}
x^2  6x \alert{{}+9}=7\alert{{}+9}
\end{equation*}
Step 3 The left side of the equation is now the square of a binomial, namely \((x  3)^2\text{.}\) We write the left side in its square form and simplify the right side, which gives us
\begin{equation*}
(x  3)^2 =16
\end{equation*}
(You can check that this equation is equivalent to the original one; if you expand the left side and collect like terms, you will return to the original equation.)
Step 4 We can now use extraction of roots to find the solutions. Taking square roots of both sides, we get
\begin{equation*}
\begin{aligned}[t]
x  3 \amp =4 \amp\text{or}\amp\amp x  3 \amp= 4\amp \blert{\text{Solve each equation.}}\\
x \amp =7 \amp\text{or}\amp\amp x \amp= 1
\end{aligned}
\end{equation*}
The solutions are \(7\) and \(1\text{.}\)
The graph of \(y = x^2  6x  7\) is shown in at left. Note that the \(x\)intercepts of the graph are \(x = 7\) and \(x = 1\text{,}\) and the parabola is symmetric about the vertical line halfway between the intercepts, at \(x = 3\text{.}\)
We can also solve \(x^2  6x  7 = 0\) by factoring instead of completing the square. Of course, we get the same solutions by either method. In Example 6.21, we will solve an equation that cannot be solved by factoring.
Example 6.21
Solve \(x^2  4x  3 = 0\) by completing the square.
Solution

First, write the equation with the constant term on the right side.
\begin{equation*}
x^2  4x{}~\underline{\hspace{2.727272727272727em}}{}=3
\end{equation*}
 Now complete the square on the left side. The coefficient of \(x\) is \(4\text{,}\) so
\begin{equation*}
p = \frac{1}{2}(4) = 2 ~~~\text{ and } ~~~ p^2 = (2)^2 = 4
\end{equation*}
We add \(4\) to both sides of our equation:
\begin{equation*}
x^2  4x \alert{{}+4}=3\alert{{}+4}
\end{equation*}

Write the left side as the square of a binomial, and combine terms on the right side:
\begin{equation*}
(x  2)^2 =7
\end{equation*}

Finally, use extraction of roots to obtain
\begin{equation*}
\begin{aligned}[t]
x  2 \amp =\sqrt{7} \amp\text{or}\amp\amp x  2 \amp= \sqrt{7}\amp \blert{\text{Solve each equation.}}\\
x \amp =2+\sqrt{7} \amp\text{or}\amp\amp x \amp=2 \sqrt{7}
\end{aligned}
\end{equation*}
The solutions are \(2+\sqrt{7}\approx 4.646\) and \(2\sqrt{7}\approx 0.646\text{.}\) The graph of \(y = x^2  4x  3\) is shown below.
Checkpoint 6.22

Follow the steps to solve by completing the square: \(x^2  1 = 3x\text{.}\)
Write the equation with the constant on the right.

Complete the square on the left:
\begin{equation*}
p = \frac{1}{2}(3) ={}\underline{\hspace{2.727272727272727em}}\text{, } ~~~{} p^2 = \underline{\hspace{2.727272727272727em}}
\end{equation*}
Add \(p^2\) to both sides.
Write the left side as a perfect square; simplify the right side.
Solve by extracting roots.
Find approximations to two decimal places for the solutions.

Graph the parabola \(y = x^2  3x  1\) in the window
\begin{equation*}
\begin{aligned}[t]
\text{Xmin} \amp = 4.7 \amp\amp \text{Xmax} = 4.7\\
\text{Ymin} \amp = 5 \amp\amp \text{Ymax} = 5
\end{aligned}
\end{equation*}
Answer
\(x=\dfrac{3}{2}\pm\sqrt{\frac{13}{4}} \)
\(x\approx 0.30\) or \(x\approx 3.30\)
Subsection The General Case
Our method for completing the square works only if the coefficient of \(x^2\) is \(1\text{.}\) If we want to solve a quadratic equation whose lead coefficient is not \(1\text{,}\) we first divide each term of the equation by the lead coefficient.
Example 6.23
Solve \(~~2x^2  6x  5 = 0\text{.}\)
Solution

Because the coefficient of \(x^2\) is \(2\text{,}\) we must divide each term of the equation by \(2\text{.}\)
\begin{equation*}
x^2  3x  \frac{5}{2}= 0
\end{equation*}
Now we proceed as before. Rewrite the equation with the constant on the right side.
\begin{equation*}
x^2  3x ~\underline{\hspace{2.727272727272727em}} = \frac{5}{2}
\end{equation*}

Complete the square:
\begin{equation*}
p = \frac{1}{2}(3) = \frac{3}{2} ~~~\text{ and } ~~~ p^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}
\end{equation*}
Add \(\frac{9}{4}\) to both sides of our equation:
\begin{equation*}
x^2  3x \alert{{}+\frac{9}{4}}=\frac{5}{2}\alert{{}+\frac{9}{4}}
\end{equation*}

Rewrite the left side as the square of a binomial and simplify the right side to get
\begin{equation*}
\left(x  \frac{3}{2}\right)^2 =\frac{19}{4}
\end{equation*}

Finally, extract roots and solve each equation for \(x\text{.}\)
\begin{equation*}
x  \frac{3}{2} =\sqrt{\frac{19}{4}} ~~~\text{ or }~~~ x  \frac{3}{2} = \sqrt{\frac{19}{4}}
\end{equation*}
The solutions are \(\dfrac{3}{2}+\sqrt{\dfrac{19}{4}}\) and \(\dfrac{3}{2}\sqrt{\dfrac{19}{4}}\text{.}\)
Using a calculator, we can find decimal approximations for the solutions: \(3.679\) and \(0.679\text{.}\)
Checkpoint 6.25

Follow the steps to solve by completing the square:
\begin{equation*}
4x^2  3665 = 0\text{.}
\end{equation*}
Divide each term by \(4\text{.}\) Write the equation with the constant on the right.

Complete the square on the left:
\begin{equation*}
p = \frac{1}{2}(9) ={}\underline{\hspace{2.727272727272727em}}\text{, } ~~~{} p^2 = \underline{\hspace{2.727272727272727em}}
\end{equation*}
Add \(p^2\) to both sides.
Write the left side as a perfect square; simplify the right side.
Solve by extracting roots.

Graph \(y = 4x^2  36x  65\) in the window
\begin{equation*}
\begin{aligned}[t]
\text{Xmin} \amp = 9.4 \amp\amp \text{Xmax} = 0\\
\text{Ymin} \amp = 10 \amp\amp \text{Ymax} = 20
\end{aligned}
\end{equation*}
Answer\(x=\dfrac{13}{2} \text{,}\) \(~x=\dfrac{5}{2} \)
Here is a summary of the steps for solving quadratic equations by completing the square.
To Solve a Quadratic Equation by Completing the Square
Write the equation in standard form.
Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides.

Complete the square on the left side:
 Multiply the coefficient of the firstdegree term by onehalf, then square the result.
 Add the value obtained in (a) to both sides of the equation.
Write the left side of the equation as the square of a binomial. Simplify the right side.
Use extraction of roots to finish the solution.
Subsection Quadratic 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\text{, }~~~ a \ne 0
\end{equation*}
and obtain a formula for the solutions of any quadratic equation.
The Quadratic Formula
The solutions of the equation \(ax^2 + bx + c = 0\text{, }~~~ a \ne 0\text{,}\) are
\begin{equation*}
\blert{x=\frac{b \pm \sqrt{b^2  4ac}}{2a}}
\end{equation*}
This formula expresses the solutions of a quadratic equation in terms of its coefficients. (The proof of the formula is considered in the Homework problems.) The symbol \(\pm\text{,}\) read plus or minus, is used to combine the two equations
\begin{equation*}
x =\dfrac{b + \sqrt{b^2  4ac}}{2a} ~~~~ \text{and} ~~~~x =\dfrac{b  \sqrt{b^2  4ac}}{2a}
\end{equation*}
into a single equation.
To solve a quadratic equation using the quadratic formula, all we have to do is substitute the coefficients \(a\text{,}\) \(b\text{,}\) and \(c\) into the formula.
Example 6.26
Solve \(~~2x^2 + 1 = 4x\text{.}\)
Solution
Write the equation in standard form as
\begin{equation*}
2x^2  4x + 1 = 0
\end{equation*}
Substitute \(\alert{2}\) for \(a\text{,}\) \(\alert{4}\) for \(b\text{,}\) and \(\alert{1}\) for \(c\) into the quadratic formula, then simplify.
\begin{equation*}
\begin{aligned}[t]
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{aligned}
\end{equation*}
Using a calculator, we find that the solutions are approximately \(1.707\) and \(0.293\text{.}\)
We can also verify that the \(x\)intercepts of the graph of \(y = 2x^2  4x + 1\) are approximately \(1.707\) and \(0.293\text{,}\) as shown below.
Checkpoint 6.27
Use the quadratic formula to solve \(~~x^2  3x = 1\text{.}\)
Answer\(x=\dfrac{3\pm \sqrt{13}}{2} \)
Note that the solutions to this equation are the same as the solutions we found in Checkpoint 6.22.
Subsection 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.
Example 6.28
The owners of a daycare 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{equation*}
x(300  3x) = 6000
\end{equation*}
This is a quadratic equation. In standard form,
\begin{equation*}
\begin{aligned}[t]
3x^2  300x + 6000 \amp= 0\amp\amp \blert{\text{Divide each term by 3.}}\\
x^2  100x + 2000 \amp= 0
\end{aligned}
\end{equation*}
The left side cannot be factored, so we use the quadratic formula with \(a = \alert{1}\text{,}\) \(b = \alert{100}\text{,}\) and \(c = \alert{2000}\text{.}\)
\begin{equation*}
\begin{aligned}[t]
x \amp=\frac{(\alert{100}) \pm\sqrt{(\alert{100})^2  4(\alert{1})(\alert{2000})}}{2(\alert{1})}\\
\amp= \frac{100 \pm\sqrt{2000}}{2}\approx \frac{100 \pm 44.7}{2}
\end{aligned}
\end{equation*}
Simplifying the last fraction, we find that \(x \approx 72.35\) or \(x\approx 27.65\text{.}\) Both values give solutions to the problem.
 If the width of the play area is \(72.35\) feet, then the length is \(300  3(72.35)\text{,}\) or \(82.95\) feet.
 If the width is \(27.65\) feet, the length is \(300  3(27.65)\text{,}\) or \(217.05\) feet.
Checkpoint 6.29
In Investigation 6.1, we considered the height of a baseball, given by the equation
\begin{equation*}
h = 16t^2 + 64t + 4
\end{equation*}
Find two times when the ball is at a height of \(20\) feet. Give your answers to two decimal places.
Answer\(0.27\) sec, \(3.73\) sec
Sometimes it is useful to solve a quadratic equation for one variable in terms of the others.
Example 6.30
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 \(c = \alert{y  2}\text{.}\) Substitute these expressions into the quadratic formula.
\begin{equation*}
\begin{aligned}[t]
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{aligned}
\end{equation*}
Checkpoint 6.31
Solve \(~~2x^2 + kx + k^2 = 1~~\) for \(x\) in terms of \(k\text{.}\)
Answer\(x=\dfrac{k\pm\sqrt{87k^2}}{4} \)
Subsection Section Summary
¶
Subsubsection Vocabulary
Look up the definitions of new terms in the Glossary.
Quadratic trinomial
Complete the square
Quadratic formula
Subsubsection CONCEPTS

The square of the binomial is a quadratic trinomial,
\begin{equation*}
(x + p)^2 = x^2 + 2px + p^2
\end{equation*}
To Solve a Quadratic Equation by Completing the Square
Write the equation in standard form.
Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides.

Complete the square on the left side:
 Multiply the coefficient of the firstdegree term by onehalf, then square the result.
 Add the value obtained in (a) to both sides of the equation.
Write the left side of the equation as the square of a binomial. Simplify the right side.
Use extraction of roots to finish the solution.
The Quadratic Formula
The solutions of the equation \(ax^2 + bx + c = 0\text{, }~~~ a \ne 0\text{,}\) are
\begin{equation*}
\blert{x=\frac{b \pm \sqrt{b^2  4ac}}{2a}}
\end{equation*}
We have four methods for solving quadratic equations: extracting of roots, factoring, completing the square, and using the quadratic formula. The first two methods are faster, but they don't work on all equations. The last two methods work on any quadratic equation.
Subsubsection STUDY QUESTIONS
Name four algebraic methods for solving a quadratic equation.
Give an example of a quadratic trinomial that is the square of a binomial.
What number must be added to \(x^2  26x\) to make it the square of a binomial?
After completing the square, how do we finish solving the quadratic equation?
What is the first step in solving the equation \(2x^2  6x = 5\) by completing the square?
Subsubsection SKILLS
Practice each skill in the Homework problems listed.
Solve quadratic equations by completing the square: #3–24
Solve quadratic equations by using the quadratic formula: #27–36
Solve problems by writing and solving quadratic equations: #37–44
Solve formulas: #45–64
Subsection Homework 6.2
¶
For Problems 12, complete the square and write the result as the square of a binomial.
1
\(x^2+8x\)
\(x^27x\)
\(x^2+\dfrac{3}{2}x \)
\(x^2\dfrac{4}{5}x \)
2
\(x^214x\)
\(x^2+3x\)
\(x^2\dfrac{5}{2}x \)
\(x^2+\dfrac{2}{3}x \)
For Problems 318, solve by completing the square.
3
\(x^2  2x + 1 = 0\)
4
\(x^2 +4x + 4 = 0\)
5
\(x^2 +9x + 20 = 0 \)
6
\(x^2  x  20 = 0\)
7
\(x^2 = 3  3x\)
8
\(x^2 = 5  5x\)
9
\(2x^2 + 4x  3 = 0\)
10
\(3x^2 + 12x + 2 = 0\)
11
\(3x^2 + x = 4\)
12
\(4x^2 + 6x = 3\)
13
\(4x^2  3 = 2x\)
14
\(2x^2  5 = 3x\)
15
\(3x^2  x  4 = 0\)
16
\(2x^2  x  3 = 0\)
17
\(5x^2 + 8x = 4\)
18
\(9x^2  12x  5 = 0\)
For Problems 1924, solve by completing the square. Your answers will involve \(a\text{,}\) \(b\text{,}\) or \(c\text{.}\)
19
\(x^2 + 2x + c = 0 \)
20
\(x^2  4x + c = 0 \)
21
\(x^2 + bx + 1 = 0 \)
22
\(x^2 + bx  4 = 0 \)
23
\(ax^2 + 2x  4 = 0 \)
24
\(ax^2  4x + 9 = 0 \)
25
Write an expression for the area of the square in the figure.
Express the area as a polynomial.
Divide the square into four pieces whose areas are given by the terms of your answer to part (b).
26
Write an expression for the area of the shaded region in the figure.
Express the area in factored form.
By making one cut in the shaded region, rearrange the pieces into a rectangle whose area is given by your answer to part (b).
For Problems 2336, solve using the quadratic formula. Round your answers to three decimal places.
27
\(x^2  x  1 = 0\)
28
\(x^2 + x  1 = 0\)
29
\(y^2 + 2y = 5\)
30
\(y^2  4y = 4\)
31
\(3z^2 = 4.2z + 1.5\)
32
\(2z^2 = 7.5z  6.3\)
33
\(0=x^2 \dfrac{5}{3}x+\dfrac{1}{3} \)
34
\(0 = x^2 + \dfrac{5}{2}x\dfrac{1}{2} \)
35
\(5.2z^2 + 176z + 1218 = 0 \)
36
\(15z^2  18z  2750 = 0 \)
37
A car traveling at \(s\) miles per hour on a dry road surface requires approximately \(d\) feet to stop, where \(d\) is given by the function
\begin{equation*}
d = f(s) = \dfrac{s^2}{24}+\dfrac{s}{2}
\end{equation*}
Make a table showing the stopping distance, \(d\text{,}\) for speeds of \(10\text{,}\) \(20\text{,}\) \(\ldots\) , \(100\) miles per hour. (Use the Table feature of your calculator.)
Graph the function for \(d\) in terms of \(s\text{.}\) Use your table values to help you choose appropriate window settings.
Write and solve an equation to answer the question: If a car must be able to stop in \(50\) feet, what is the maximum safe speed it can travel? Verify your answer on your graph.
38
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 function
\begin{equation*}
d = f(s) = \dfrac{s^2}{12}+\dfrac{s}{2}
\end{equation*}
Make a table showing the stopping distance, \(d\text{,}\) for speeds of \(10\text{,}\) \(20\text{,}\) \(\ldots\) , \(100\) miles per hour. (Use the Table feature of your calculator.)
Graph the function 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.
39
A skydiver jumps out of an airplane at \(11,000\) feet. While she is in freefall, her altitude in feet \(t\) seconds after jumping is given by the function
\begin{equation*}
h = f(t) = 16t^2  16t + 11,000
\end{equation*}
Make a table of values showing the skydiver's altitude at \(5\)second intervals after she jumps from the airplane. (Use the Table feature of your calculator.)
Graph the function. Use your table of values to choose appropriate window settings.
If the skydiver must open her parachute at an altitude of \(1000\) feet, how long can she freefall? Write and solve an equation to find the answer.
If the skydiver drops a marker just before she opens her parachute, how long will it take the marker to hit the ground? (Hint: The marker continues to fall according to the equation given above.)
Find points on your graph that correspond to your answers to parts (c) and (d).
40
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 the function
\begin{equation*}
h = f(t) = 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 airplane. (Use the Table feature of your calculator.)
Graph the function. 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).
41
A dog trainer has \(100\) meters of chain link fence. She wants to enclose \(250\) square meters in three pens of equal size, as shown in the figure.
Let \(l\) and \(w\) represent the length and width, respectively, of the entire area. Write an equation about the amount of chain link fence.
Solve your equation for \(l\) in terms \(w\text{.}\)
Write and solve an equation in \(w\) for the total area enclosed.
Find the dimensions of each pen.
42
An architect is planning to include a rectangular window topped by a semicircle in his plans for a new house, as shown in the figure. In order to admit enough light, the window should have an area of \(120\) square feet. The architect wants the rectangular portion of the window to be \(2\) feet wider than it is tall.
Let \(x\) stand for the horizontal width of the window. Write expressions for the height of the rectangular portion and for the radius of the semicircular portion.
Write an expression for the total area of the window.
Write and solve an equation to find the width and overall height of the window.
43
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.
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? (You will need to use the Pythagorean theorem and the fact that the radius of the Earth is \(3960\) miles. Do not 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?
44
If the radius of the Earth is \(6370\) kilometers, how far can you see from an airplane at an altitude of \(10,000\) meters? (Hint: See Problem 43.)
b. How high would the airplane have to be in order for you to see a distance of \(10\) kilometers?
For Problems 4552, use the quadratic formula to solve each equation for the indicated variable.
45
\(A = 2w^2 + 4lw ,~~\) for \(w\)
46
\(A = \pi r^2 + \pi rs ,~~\) for \(r\)
47
\(h = 4t  16t^2 ,~~\) for \(t\)
48
\(P = IE  RI^2 ,~~\) for \(I\)
49
\(s=vt\dfrac{1}{2} at^2 ,~~\) for \(t\)
50
\(S=\dfrac{n^2+n}{2} ,~~ \) for \(n\)
51
\(3x^2 + xy + y^2 = 2 ,~~\) for \(y\)
52
\(y^2  3xy + x^2 = 3 ,~~ \) for \(x\)
For Problems 5360, solve for \(y\) in terms of \(x\text{.}\) Use whichever method of solution seems easiest.
53
\(x^2 y  y^2 = 0\)
54
\(x^2 y^2  y = 0\)
55
\((2y + 3x)^2 = 9\)
56
\((3y  2x)^2 = 4\)
57
\(4x^2  9y^2 = 36\)
58
\(9x^2 + 4y^2 = 36\)
59
\(4x^2  25y^2 = 0\)
60
\((2x  5y)^2 = 0\)
For Problems 6166, solve the formula for the indicated variable.
61
\(V = \pi (r  3)^2h , ~~ \) for \(r\)
62
\(A = P(1 + r )^2 , ~~ \) for \(r\)
63
\(E = \dfrac{1}{2} mv^2 + mgh , ~~ \) for \(v\)
64
\(h = \dfrac{1}{2}gt^2 + dl , ~~ \) for \(t\)
65
\(V = 2(s^2 + t^2)w , ~~ \) for \(t\)
66
\(V = \pi(r ^2 + R^2)h , ~~ \) for \(R\)
67
What is the sum of the two solutions of the quadratic equation \(ax^2 + bx + c = 0\text{?}\)
HintThe two solutions are given by the quadratic formula.
68
What is the product of the two solutions of the quadratic equation \(ax^2 + bx + c = 0\text{?}\)
HintDo not try to multiply the two solutions given by the quadratic formula! Think about the factored form of the equation
In Problems 69 and 70, we prove the quadratic formula.
69
Complete the square to find the solutions of the equation \(x^2 + bx + c = 0\text{.}\) (Your answers will be expressions in \(b\) and \(c\text{.}\))
70
Complete the square to find the solutions of the equation \(ax^2 + bx + c = 0\text{.}\) (Your answers will be expressions in \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\))