Elementary Algebra 2e · free preview

Chapter 10 Review and Practice Test

Learning objectives: Review the key ideas of this chapter.; Apply the chapter's methods to review exercises.; Check your understanding with a practice test.

1. Learning objectives

Learning objectives

Learning objectives

  • Review the key ideas of this chapter.
  • Apply the chapter's methods to review exercises.
  • Check your understanding with a practice test.

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2. Worked example

Worked example

axis of symmetry

completing the square

Completing the square is a method used to solve quadratic equations.

consecutive even integers

consecutive odd integers

discriminant

parabola

The graph of a quadratic equation in two variables is a parabola.

quadratic equation

quadratic equation in two variables

Square Root Property

vertex

vertex

x

x

y

y

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Key Concepts

  • Square Root PropertyIf x2=k{x}^{2}=k, and k≥0k\ge{}0, then x=k or x=−kx=\sqrt{k}\,\mathrm{or}\,x=-\sqrt{k}.
  • Binomial Squares Pattern If a,ba,b are real numbers,(a+b)2=a2+2ab+b2{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}(a−b)2=a2−2ab+b2{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}
  • Complete a SquareTo complete the square of x2+bx{x}^{2}+bx: Step 1. Identify bb, the coefficient of xx. Step 2. Find (12b)2{(\frac{1}{2}b)}^{2}, the number to complete the square. Step 3. Add the (12b)2{(\frac{1}{2}b)}^{2} to x2+bx{x}^{2}+bx.
  • Quadratic Formula The solutions to a quadratic equation of the form ax2+bx+c=0,a{x}^{2}+bx+c=0,a≠0a\ne{}0 are given by the formula:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}
  • Solve a Quadratic Equation Using the Quadratic Formula To solve a quadratic equation using the Quadratic Formula. Step 1. Write the quadratic formula in standard form. Identify the a,b,ca,b,c values. Step 2. Write the quadratic formula. Then substitute in the values of a,b,c.a,b,c. Step 3. Simplify. Step 4. Check the solutions.
  • Using the Discriminant, b2−4ac{b}^{2}-4ac, to Determine the Number of Solutions of a Quadratic Equation For a quadratic equation of the form ax2+bx+c=0,a{x}^{2}+bx+c=0,a≠0,a\ne{}0, if b2−4ac>0{b}^{2}-4ac>0, the equation has 2 solutions. if b2−4ac=0{b}^{2}-4ac=0, the equation has 1 solution. if b2−4ac<0{b}^{2}-4ac<0, the equation has no real solutions.
  • To identify the most appropriate method to solve a quadratic equation: Step 1. Try Factoring first. If the quadratic factors easily this method is very quick. Step 2. Try the Square Root Property next. If the equation fits the form ax2=ka{x}^{2}=k or a(x−h)2=ka{(x-h)}^{2}=k, it can easily be solved by using the Square Root Property. Step 3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.
  • Area of a Triangle For a triangle with base, bb, and height, hh, the area, AA, is given by the formula: A=12bhA=\frac{1}{2}bh
  • Pythagorean Theorem In any right triangle, where aa and bb are the lengths of the legs, and cc is the length of the hypothenuse, a2+b2=c2{a}^{2}+{b}^{2}={c}^{2}
  • Projectile motion The height in feet, hh, of an object shot upwards into the air with initial velocity, v0{v}_{0}, after tt seconds can be modeled by the formula:
h=−16t2+v0th=-16{t}^{2}+{v}_{0}t
  • The graph of every quadratic equation is a parabola.
  • Parabola Orientation For the quadratic equation y=ax2+bx+cy=a{x}^{2}+bx+c, if a>0a>0, the parabola opens upward. a<0a<0, the parabola opens downward.
  • Axis of Symmetry and Vertex of a Parabola For a parabola with equation y=ax2+bx+cy=a{x}^{2}+bx+c: The axis of symmetry of a parabola is the line x=−b2ax=-\frac{b}{2a}. The vertex is on the axis of symmetry, so its x-coordinate is −b2a-\frac{b}{2a}. To find the y-coordinate of the vertex we substitute x=−b2ax=-\frac{b}{2a} into the quadratic equation.
  • Find the Intercepts of a Parabola To find the intercepts of a parabola with equation y=ax2+bx+cy=a{x}^{2}+bx+c:y−interceptx−interceptsLet x=0 andsolvefor y.Let y=0 andsolvefor x.\begin{array}{rccr} y\mathrm{-intercept} & & & x\mathrm{-intercepts} \\ \mathrm{Let}\,x=0\,\mathrm{and solve for}\,y. & & & \mathrm{Let}\,y=0\,\mathrm{and solve for}\,x. \\ \end{array}
  • To Graph a Quadratic Equation in Two Variables Step 1. Write the quadratic equation with yy on one side. Step 2. Determine whether the parabola opens upward or downward. Step 3. Find the axis of symmetry. Step 4. Find the vertex. Step 5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry. Step 6. Find the x-intercepts. Step 7. Graph the parabola.
  • Minimum or Maximum Values of a Quadratic Equation The y-coordinate of the vertex of the graph of a quadratic equation is the minimum value of the quadratic equation if the parabola opens upward. maximum value of the quadratic equation if the parabola opens downward.

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Review Exercises

In the following exercises, solve using the Square Root Property.

In the following exercises, solve using the Square Root Property.

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

In the following exercises, solve by completing the square.

In the following exercises, solve by using the Quadratic Formula.

In the following exercises, determine the number of solutions to each quadratic equation.

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation.

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

In the following exercises, graph by plotting point.

In the following exercises, determine if the following parabolas open up or down.

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

In the following exercises, find the x- and y-intercepts.

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

In the following exercises, find the minimum or maximum value.

In the following exercises, solve. Rounding answers to the nearest tenth.

##

Solve the following quadratic equations. Use any method.

Use the discriminant to determine the number of solutions of each quadratic equation.

Solve by factoring, the Square Root Property, or the Quadratic Formula.

For each parabola, find ⓐ which ways it opens, ⓑ the axis of symmetry, ⓒ the vertex, ⓓ the x- and y-intercepts, and ⓔ the maximum or minimum value.

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.

Solve.

3. Key Terms

Key Terms

axis of symmetry

completing the square

Completing the square is a method used to solve quadratic equations.

consecutive even integers

consecutive odd integers

discriminant

parabola

The graph of a quadratic equation in two variables is a parabola.

quadratic equation

quadratic equation in two variables

Square Root Property

vertex

vertex

x

x

y

y

##

Key Concepts

  • Square Root PropertyIf x2=k{x}^{2}=k, and k≥0k\ge{}0, then x=k or x=−kx=\sqrt{k}\,\mathrm{or}\,x=-\sqrt{k}.
  • Binomial Squares Pattern If a,ba,b are real numbers,(a+b)2=a2+2ab+b2{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}(a−b)2=a2−2ab+b2{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}
  • Complete a SquareTo complete the square of x2+bx{x}^{2}+bx: Step 1. Identify bb, the coefficient of xx. Step 2. Find (12b)2{(\frac{1}{2}b)}^{2}, the number to complete the square. Step 3. Add the (12b)2{(\frac{1}{2}b)}^{2} to x2+bx{x}^{2}+bx.
  • Quadratic Formula The solutions to a quadratic equation of the form ax2+bx+c=0,a{x}^{2}+bx+c=0,a≠0a\ne{}0 are given by the formula:
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}
  • Solve a Quadratic Equation Using the Quadratic Formula To solve a quadratic equation using the Quadratic Formula. Step 1. Write the quadratic formula in standard form. Identify the a,b,ca,b,c values. Step 2. Write the quadratic formula. Then substitute in the values of a,b,c.a,b,c. Step 3. Simplify. Step 4. Check the solutions.
  • Using the Discriminant, b2−4ac{b}^{2}-4ac, to Determine the Number of Solutions of a Quadratic Equation For a quadratic equation of the form ax2+bx+c=0,a{x}^{2}+bx+c=0,a≠0,a\ne{}0, if b2−4ac>0{b}^{2}-4ac>0, the equation has 2 solutions. if b2−4ac=0{b}^{2}-4ac=0, the equation has 1 solution. if b2−4ac<0{b}^{2}-4ac<0, the equation has no real solutions.
  • To identify the most appropriate method to solve a quadratic equation: Step 1. Try Factoring first. If the quadratic factors easily this method is very quick. Step 2. Try the Square Root Property next. If the equation fits the form ax2=ka{x}^{2}=k or a(x−h)2=ka{(x-h)}^{2}=k, it can easily be solved by using the Square Root Property. Step 3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.
  • Area of a Triangle For a triangle with base, bb, and height, hh, the area, AA, is given by the formula: A=12bhA=\frac{1}{2}bh
  • Pythagorean Theorem In any right triangle, where aa and bb are the lengths of the legs, and cc is the length of the hypothenuse, a2+b2=c2{a}^{2}+{b}^{2}={c}^{2}
  • Projectile motion The height in feet, hh, of an object shot upwards into the air with initial velocity, v0{v}_{0}, after tt seconds can be modeled by the formula:
h=−16t2+v0th=-16{t}^{2}+{v}_{0}t
  • The graph of every quadratic equation is a parabola.
  • Parabola Orientation For the quadratic equation y=ax2+bx+cy=a{x}^{2}+bx+c, if a>0a>0, the parabola opens upward. a<0a<0, the parabola opens downward.
  • Axis of Symmetry and Vertex of a Parabola For a parabola with equation y=ax2+bx+cy=a{x}^{2}+bx+c: The axis of symmetry of a parabola is the line x=−b2ax=-\frac{b}{2a}. The vertex is on the axis of symmetry, so its x-coordinate is −b2a-\frac{b}{2a}. To find the y-coordinate of the vertex we substitute x=−b2ax=-\frac{b}{2a} into the quadratic equation.
  • Find the Intercepts of a Parabola To find the intercepts of a parabola with equation y=ax2+bx+cy=a{x}^{2}+bx+c:y−interceptx−interceptsLet x=0 andsolvefor y.Let y=0 andsolvefor x.\begin{array}{rccr} y\mathrm{-intercept} & & & x\mathrm{-intercepts} \\ \mathrm{Let}\,x=0\,\mathrm{and solve for}\,y. & & & \mathrm{Let}\,y=0\,\mathrm{and solve for}\,x. \\ \end{array}
  • To Graph a Quadratic Equation in Two Variables Step 1. Write the quadratic equation with yy on one side. Step 2. Determine whether the parabola opens upward or downward. Step 3. Find the axis of symmetry. Step 4. Find the vertex. Step 5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry. Step 6. Find the x-intercepts. Step 7. Graph the parabola.
  • Minimum or Maximum Values of a Quadratic Equation The y-coordinate of the vertex of the graph of a quadratic equation is the minimum value of the quadratic equation if the parabola opens upward. maximum value of the quadratic equation if the parabola opens downward.

##

Review Exercises

In the following exercises, solve using the Square Root Property.

In the following exercises, solve using the Square Root Property.

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

In the following exercises, solve by completing the square.

In the following exercises, solve by using the Quadratic Formula.

In the following exercises, determine the number of solutions to each quadratic equation.

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation.

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

In the following exercises, graph by plotting point.

In the following exercises, determine if the following parabolas open up or down.

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

In the following exercises, find the x- and y-intercepts.

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

In the following exercises, find the minimum or maximum value.

In the following exercises, solve. Rounding answers to the nearest tenth.

##

Solve the following quadratic equations. Use any method.

Use the discriminant to determine the number of solutions of each quadratic equation.

Solve by factoring, the Square Root Property, or the Quadratic Formula.

For each parabola, find ⓐ which ways it opens, ⓑ the axis of symmetry, ⓒ the vertex, ⓓ the x- and y-intercepts, and ⓔ the maximum or minimum value.

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.

Solve.

4. Practice 1

  1. x2=100{x}^{2}=100
  1. m2−40=0{m}^{2}-40=0
  1. 4a2=1004{a}^{2}=100
  1. r2+32=0{r}^{2}+32=0

5. Practice 2

  1. 43v2+4=28\frac{4}{3}{v}^{2}+4=28
  1. 5c2+3=195{c}^{2}+3=19
  1. (p−5)2+3=19{(p-5)}^{2}+3=19
  1. (u+1)2=45{(u+1)}^{2}=45

6. Practice 3

  1. (x−14)2=316{(x-\frac{1}{4})}^{2}=\frac{3}{16}
  1. (m−7)2+6=30{(m-7)}^{2}+6=30
  1. (5c+3)2=−20{(5c+3)}^{2}=-20
  1. m2−6m+9=48{m}^{2}-6m+9=48

7. Practice 4

  1. 64a2+48a+9=8164{a}^{2}+48a+9=81
  1. x2+22x{x}^{2}+22x
  1. m2−8m{m}^{2}-8m
  1. a2−3a{a}^{2}-3a

8. Practice 5

  1. p2+45p{p}^{2}+\frac{4}{5}p
  1. c2+20c=21{c}^{2}+20c=21
  1. x2−4x=32{x}^{2}-4x=32
  1. r2+6r=−100{r}^{2}+6r=-100

9. Practice 6

  1. v2−14v=−31{v}^{2}-14v=-31
  1. m2+10m−4=−13{m}^{2}+10m-4=-13
  1. a2=3a+8{a}^{2}=3a+8
  1. (u+8)(u+4)=14(u+8)(u+4)=14

10. Practice 7

  1. 3p2−18p+15=153{p}^{2}-18p+15=15
  1. 4y2−6y=44{y}^{2}-6y=4
  1. 3c2+2c=93{c}^{2}+2c=9
  1. 4x2−5x+1=04{x}^{2}-5x+1=0

11. Practice 8

  1. r2−r−42=0{r}^{2}-r-42=0
  1. 4v2+v−5=04{v}^{2}+v-5=0
  1. 3m2+8m+2=03{m}^{2}+8m+2=0
  1. 6a2−5a+2=06{a}^{2}-5a+2=0

12. Practice 9

  1. u(u−10)+3=0u(u-10)+3=0
  1. 18p2−15p=−120\frac{1}{8}{p}^{2}-\frac{1}{5}p=-\frac{1}{20}
  1. 4c2+4c+1=04{c}^{2}+4c+1=0
  1. 1. ⓐ 9x2−6x+1=09{x}^{2}-6x+1=0
  1. ⓑ 3y2−8y+1=03{y}^{2}-8y+1=0
  1. ⓒ 7m2+12m+4=07{m}^{2}+12m+4=0
  1. ⓓ 5n2−n+1=05{n}^{2}-n+1=0

13. Practice 10

  1. 1. ⓐ 16r2−8r+1=016{r}^{2}-8r+1=0
  1. ⓑ 5t2−8t+3=95{t}^{2}-8t+3=9
  1. ⓒ 3(c+2)2=153{(c+2)}^{2}=15
  1. Find two consecutive odd numbers whose product is 323.
  1. A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.
  1. A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

14. Practice 11

  1. The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.
  1. A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula h=−16t2+v0th=-16{t}^{2}+{v}_{0}t to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.
  1. Graph y=x2−2y={x}^{2}-2
  1. y=−3x2+3x−1y=-3{x}^{2}+3x-1

15. Practice 12

  1. y=x2+8x−1y={x}^{2}+8x-1
  1. y=−x2+6x+8y=-{x}^{2}+6x+8
  1. y=x2−4x−5y={x}^{2}-4x-5
  1. y=x2−4x+10y={x}^{2}-4x+10

16. Practice 13

  1. y=16x2−8x+1y=16{x}^{2}-8x+1
  1. y=x2+8x+15y={x}^{2}+8x+15
  1. y=−x2+8x−16y=-{x}^{2}+8x-16
  1. y=x2+6x+13y={x}^{2}+6x+13

17. Practice 15

  1. Use the Quadratic Formula to solve the quadratic equation: 2m2−5m+3=02{m}^{2}-5m+3=0.
  1. 3n2+8n+3=03{n}^{2}+8n+3=0
  1. x(x+3)+12=0x(x+3)+12=0
  1. Find two consecutive even numbers whose product is 360.

18. Practice 14

  1. y=−4x2+16x−11y=-4{x}^{2}+16x-11
  1. y=7x2+14x+6y=7{x}^{2}+14x+6
  1. A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation h=−16t2+112th=-16{t}^{2}+112t to find how long it will take the ball to reach maximum height, and then find the maximum height.
  1. Use the Square Root Property to solve the quadratic equation: 3(w+5)2=273{(w+5)}^{2}=27.

19. Practice 16

  1. y=3x2+6x+8y=3{x}^{2}+6x+8
  1. y=x2+10x+24y={x}^{2}+10x+24
  1. y=−x2−8x+16y=-{x}^{2}-8x+16
  1. y=16x2+24x+9y=16{x}^{2}+24x+9

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