9.E: Review Exercises and Sample Exam
- Page ID
- 21627
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Review Exercises
Solve by extracting the roots.
- \(x^{2}−16=0\)
- \(y^{2}=94\)
- \(x^{2}−27=0\)
- \(x^{2}+27=0\)
- \(3y^{2}−25=0\)
- \(9x^{2}−2=0\)
- \((x−5)^{2}−9=0\)
- \((2x−1)^{2}−1=0\)
- \(16(x−6)^{2}−3=0\)
- \(2(x+3)^{2}−5=0\)
- \((x+3)(x−2)=x+12\)
- \((x+2)(5x−1)=9x−1\)
- Answer
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1. \(±16\)
3. \(±3\sqrt{3}\)
5. \(±\frac{5 \sqrt{3}}{3}\)
7. \(2, 8\)
9. \(6±\frac{\sqrt{3}}{4}\)
11. \(±3\sqrt{2}\)
Find a quadratic equation in standard form with the given solutions.
- \(\pm\sqrt{2}\)
- \(\pm2\sqrt{5}\)
- Answer
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1. \(x^{2}-2=0\)
Complete the square.
- \(x^{2}-6x+?=(x-?)^{2}\)
- \(x^{2}-x+?=(x-?)^{2}\)
- Answer
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1. \(x^{2}-6x+9=(x-3)^{2}\)
Solve by completing the square.
- \(x^{2}−12x+1=0\)
- \(x^{2}+8x+3=0\)
- \(y^{2}−4y−14=0\)
- \(y^{2}−2y−74=0\)
- \(x^{2}+5x−1=0\)
- \(x^{2}−7x−2=0\)
- \(2x^{2}+x−3=0\)
- \(5x^{2}+9x−2=0\)
- \(2x^{2}−16x+5=0\)
- \(3x^{2}−6x+1=0\)
- \(2y^{2}+10y+1=0\)
- \(5y^{2}+y−3=0\)
- \(x(x+9)=5x+8\)
- \((2x+5)(x+2)=8x+7\)
- Answer
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1. \(6±\sqrt{35}\)
3. \(2±3\sqrt{2}\)
5. \(\frac{-5±\sqrt{29}}{2}\)
7. \(\frac{−3}{2}, 1\)
9. \(\frac{8±3\sqrt{6}}{2}\)
11. \(\frac{-5±\sqrt{23}}{2}\)
13. \(−2±2\sqrt{3}\)
Identify the coefficients a, b, and c used in the quadratic formula. Do not solve.
- \(x^{2}−x+4=0\)
- \(−x^{2}+5x−14=0\)
- \(x^{2}−5=0\)
- \(6x^{2}+x=0\)
- Answer
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1. \(a=1, b=−1,\) and \(c=4\)
3. \(a=1, b=0,\) and \(c=−5\)
Use the quadratic formula to solve the following.
- \(x^{2}−6x+6=0\)
- \(x^{2}+10x+23=0\)
- \(3y^{2}−y−1=0\)
- \(2y^{2}−3y+5=0\)
- \(5x^{2}−36=0\)
- \(7x^{2}+2x=0\)
- \(−x^{2}+5x+1=0\)
- \(−4x^{2}−2x+1=0\)
- \(t^{2}−12t−288=0\)
- \(t^{2}−44t+484=0\)
- \((x−3)^{2}−2x=47\)
- \(9x(x+1)−5=3x\)
- Answer
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1. \(3±\sqrt{3}\)
3. \(\frac{1±\sqrt{13}}{6}\)
5. \(±\frac{6\sqrt{5}}{5}\)
7. \(\frac{5±\sqrt{29}}{2}\)
9. \(−12, 24 \)
11. \(4±3\sqrt{6}\)
Use the discriminant to determine the number and type of solutions.
- \(−x^{2}+5x+1=0\)
- \(−x^{2}+x−1=0\)
- \(4x^{2}−4x+1=0\)
- \(9x^{2}−4=0\)
- Answer
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1. Two real solutions
3. One real solution
Solve using any method.
- \(x^{2}+4x−60=0\)
- \(9x^{2}+7x=0\)
- \(25t^{2}−1=0\)
- \(t^{2}+16=0\)
- \(x^{2}−x−3=0\)
- \(9x^{2}+12x+1=0\)
- \(4(x−1)^{2}−27=0\)
- \((3x+5)^{2}−4=0\)
- \((x−2)(x+3)=6\)
- \(x(x−5)=12\)
- \((x+1)(x−8)+28=3x\)
- \((9x−2)(x+4)=28x−9\)
- Answer
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1. \(−10, 6\)
3. \(±\frac{1}{5}\)
5. \(\frac{1±\sqrt{13}}{2}\)
7. \(1 ± \frac{3 \sqrt{3}}{2}\)
9. \(−4, 3\)
11. \(5±\sqrt{5}\)
Set up an algebraic equation and use it to solve the following.
- The length of a rectangle is 2 inches less than twice the width. If the area measures 25 square inches, then find the dimensions of the rectangle. Round off to the nearest hundredth.
- An 18-foot ladder leaning against a building reaches a height of 17 feet. How far is the base of the ladder from the wall? Round to the nearest tenth of a foot.
- The value in dollars of a new car is modeled by the function \(V(t)=125t^{2}−3,000t+22,000\), where t represents the number of years since it was purchased. Determine the age of the car when its value is $22,000.
- The height in feet reached by a baseball tossed upward at a speed of 48 feet/second from the ground is given by the function \(h(t)=−16t^{2}+48t\), where t represents time in seconds. At what time will the baseball reach a height of 16 feet?
- Answer
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1. Length: 6.14 inches; width: 4.07 inches
3. It is worth $22,000 new and when it is 24 years old.
Determine the x- and y-intercepts.
- \(y=2x^{2}+5x−3\)
- \(y=x^{2}−12\)
- \(y=5x^{2}−x+2\)
- \(y=−x^{2}+10x−25\)
- Answer
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1. x-intercepts: \((−3, 0), (\frac{1}{2}, 0)\); y-intercept: \((0, −3)\)
3. x-intercepts: none; y-intercept: \((0, 2)\)
Find the vertex and the line of symmetry.
- \(y=x^{2}−6x+1\)
- \(y=−x^{2}+8x−1\)
- \(y=x^{2}+3x−1\)
- \(y=9x^{2}−1\)
- Answer
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1. Vertex: \((3, −8)\); line of symmetry: \(x=3\)
3. Vertex: \((−\frac{3}{2}, −\frac{13}{4})\); line of symmetry: \(x=−32\)
Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
- \(y=x^{2}+8x+12\)
- \(y=−x^{2}−6x+7\)
- \(y=−2x^{2}−4\)
- \(y=x^{2}+4x\)
- \(y=4x^{2}−4x+1\)
- \(y=−2x^{2}\)
- \(y=−2x^{2}+8x−7\)
- \(y=3x^{2}−1\)
- Answer
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1.
3.
5.
7.
Determine the maximum or minimum y-value.
- \(y=x^{2}−10x+1\)
- \(y=−x^{2}+12x−1\)
- \(y=−5x^{2}+6x\)
- \(y=2x^{2}−x−1\)
- The value in dollars of a new car is modeled by the function \(V(t)=125t^{2}−3,000t+22,000\), where t represents the number of years since it was purchased. Determine the age of the car when its value is at a minimum.
- The height in feet reached by a baseball tossed upward at a speed of 48 feet/second from the ground is given by the function \(h(t)=−16t^{2}+48t\), where t represents time in seconds. What is the maximum height of the baseball?
- Answer
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1. Minimum: \(y = −24\)
3. Maximum: \(y = \frac{9}{5}\)
5. The car will have a minimum value 12 years after it is purchased.
Rewrite in terms of i.
- \(\sqrt{−36}\)
- \(\sqrt{−40}\)
- \(\sqrt{−\frac{8}{25}}\)
- -\(\sqrt{−19}\)
- Answer
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1. 6i
3. \(\frac{2 \sqrt{2} i}{5}\)
Perform the operations.
- \((2−5i)+(3+4i)\)
- \((6−7i)−(12−3i)\)
- \((2−3i)(5+i)\)
- \(4−i^{2}−3i\)
- Answer
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1. \(5−i\)
3. \(13−13i\)
Solve.
- \(9x^{2}+25=0\)
- \(3x^{2}+1=0\)
- \(y^{2}−y+5=0\)
- \(y^{2}+2y+4\)
- \(4x(x+2)+5=8x\)
- \(2(x+2)(x+3)=3(x^{2}+13)\)
- Answer
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1. \(\pm\frac{3}{3}\)
3. \(\frac{1}{2}\pm i \frac{\sqrt{19}}{2}\)
5. \(\pm i \frac{\sqrt{5}}{2}\)
Sample Exam
Solve by extracting the roots.
- \(4x^{2}−9=0\)
- \((4x+1)^{2}−5=0\)
- Answer
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1. \(\pm\frac{3}{2}\)
Solve by completing the square.
- \(x^{2}+10x+19=0\)
- \(x^{2}−x−1=0\)
- Answer
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1. \(-5\pm\sqrt{6}\)
Solve using the quadratic formula.
- \(−2x^{2}+x+3=0\)
- \(x^{2}+6x−31=0\)
- Answer
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1. \(-1, \frac{3}{2}\)
Solve using any method.
- \((5x+1)(x+1)=1\)
- \((x+5)(x−5)=65\)
- \(x(x+3)=−2\)
- \(2(x−2)^{2}−6=3x^{2}\)
- Answer
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1. \(-\frac{6}{5}, 0\)
3. \(-2, -1\)
Set up an algebraic equation and solve.
- The length of a rectangle is twice its width. If the diagonal measures \(6\sqrt{5}\) centimeters, then find the dimensions of the rectangle.
- The height in feet reached by a model rocket launched from a platform is given by the function \(h(t)=−16t^{2}+256t+3\), where t represents time in seconds after launch. At what time will the rocket reach 451 feet?
- Answer
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1. Length: 12 centimeters; width: 6 centimeters
Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
- \(y=2x^{2}−4x−6\)
- \(y=−x^{2}+4x−4\)
- \(y=4x^{2}−9\)
- \(y=x^{2}+2x−1\)
- Determine the maximum or minimum y-value: \(y=−3x^{2}+12x−15\).
- Determine the x- and y-intercepts: \(y=x^{2}+x+4\).
- Determine the domain and range: \(y=25x^{2}−10x+1\).
- The height in feet reached by a model rocket launched from a platform is given by the function \(h(t)=−16t^{2}+256t+3\), where t represents time in seconds after launch. What is the maximum height attained by the rocket.
- A bicycle manufacturing company has determined that the weekly revenue in dollars can be modeled by the formula \(R=200n−n^{2}\), where n represents the number of bicycles produced and sold. How many bicycles does the company have to produce and sell in order to maximize revenue?
- Rewrite in terms of i: \(\sqrt{−60}\).
- Divide: \(\frac{4−2i}{4+2i}\).
- Answer
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1.
3.
5. Maximum: \(y = −3\)
7. Domain: R; range: \([0,∞)\)
9. To maximize revenue, the company needs to produce and sell 100 bicycles a week.
11. \(\frac{3}{5}−i\frac{4}{5}\)
Solve.
- \(25x^{2}+3=0\)
- \(−2x^{2}+5x−1=0\)
- Answer
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2. \(\frac{5\pm\sqrt{17}}{4}\)