## Section3.5Chapter 3 Summary and Review

### Subsection3.5.1Glossary

• parabola

• extraction of roots

• compound interest

• factor

• multiplicity

• vertex

• axis of symmetry

• complete the square

### Subsection3.5.2Key Concepts

1. A quadratic equation has the standard form $$ax^2+bx=c=0\text{,}$$ where $$a,~b,~$$ and $$c$$ are constants and $$a$$ is not equal to zero.

2. The graph of a quadratic equation $$y=ax^2+bx=c$$ is called a parabola. The basic parabola is the graph of $$y=x^2\text{.}$$

3. #### Extraction of Roots.

To solve a quadratic equation of the form

\begin{equation*} ax^2+c=0 \end{equation*}
1. Isolate $$x$$ on one side of the equation.

2. Take the square root of each side.

4. Every quadratic equation has two solutions, which may be the same.

5. Solutions of a quadratic equation can be given as exact values or as decimal approximations.

6. #### Formulas for Volume and Surface Area.

1. Sphere $$~~~~~~V=\dfrac{4}{3}\pi r^3~~~~~~S=4\pi r^2$$

2. Cylinder $$~~~~~~V=\pi r^2h~~~~~~S=2\pi r^2+2 \pi rh$$

3. Cone $$~~~~~~V=\dfrac{1}{3}\pi r^2h~~~~~~S=\pi r^2+\pi rs$$

4. Square Pyramid $$~~~~~~V=\dfrac{1}{3}s^2h$$

7. The formula for interest compounded annually is

\begin{equation*} A=P(1+r)^n \end{equation*}
8. Zero-Factor Principle: The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols,

\begin{equation*} ab=0~~\text{if and only if}~~a=0~~|text{or}~~b=0~~\text{or both} \end{equation*}
9. #### To Solve a Quadratic Equation by Factoring.

1. Write the equation in standard form.

2. Factor the left side of the equation.

3. Apply the zero-factor principle: Set each factor equal to zero.

4. Solve each equation. There are two solutions (which may be equal).

10. Each solution of a quadratic equation corresponds to a non-constant factor in the factored form.

11. The value of the constant $$a$$ in the factored form of a quadratic equation does not affect the solutions.

12. The $$x$$-intercepts of the graph of $$y=ax^2+bx+c$$ are the solutions of the equation $$ax^2+bx+c=0\text{.}$$

13. #### The Graph of $$y=ax^2$$.

• The parabola opens upward if $$a \gt 0$$

• The parabola opens downward if $$a \lt 0$$

• The magnitude of $$a$$ determines how wide or narrow the parabola is.

• The vertex, the $$x$$-intercepts, and the $$y$$-intercept all coincide at the origin.

14. #### The Graph of $$y=x^2+c$$.

Compared to the graph of $$y=x^2\text{,}$$ the graph of $$y=x^2+c$$

• is shifted upward by $$c$$ units if $$c \gt 0$$

• is shifted downward by $$c$$ units if $$c \lt 0$$

15. For the graph of $$y=ax^2+bx+c\text{,}$$ the $$x$$-coordinate of the vertex is

\begin{equation*} x_v=\dfrac{-b}{2a} \end{equation*}
16. The square of a binomial is a quadratic trinomial,

\begin{equation*} (x+p)^2=x^2+2px+p^2 \end{equation*}
17. #### To Solve a Quadratic Equation by Completing the Square.

1. Write the equation in standard form.

2. Divide both sides of the equation by the coefficient of the quadratic term, and subtract the constant term from both sides.

1. Complete the square on the left side:

1. Multiply the coefficient of the first-degree term by one-half, then square the result.

2. Add the value obtained in (a) to both sides of the equation.

2. Write the left side of the equation as the square of a binomial. Simplify the right side.

3. Use extraction of roots to finish the solution.

### Exercises3.5.3Chapter 3 Review Problems

#### Exercise Group.

For Problems 1–6, solve by extraction of roots.

##### 1.

$$x^2+7=13-2x^2$$

$$\pm \sqrt{2}$$

##### 2.

$$\dfrac{2x^2}{5}-7=9$$

$$\pm \sqrt{40}$$

##### 3.

$$3(x+4)^2=60$$

$$-4 \pm \sqrt{20}$$

##### 4.

$$(7x-1)^2=15$$

$$\dfrac{1 \pm \sqrt{15}}{7}$$

##### 5.

$$(2x-5)^2=9$$

$$1,~4$$

##### 6.

$$(3.5-0.2x)^2=1.44$$

$$11.5,~23.5$$

#### Exercise Group.

For Problems 7–10, solve by factoring.

##### 7.

$$(w+1)(2w-3)=3$$

$$\dfrac{-3}{2},~2$$

##### 8.

$$6y=(y+1)^2+3$$

$$2,~2$$

##### 9.

$$4x-(x+1)(x+2)=-8$$

$$-2,~3$$

##### 10.

$$3(x+2)^2=15+12x$$

$$-1,~1$$

#### Exercise Group.

For Problems 11 and 12, write a quadratic equation with integer coefficients and with the given solutions.

##### 11.

$$\dfrac{-3}{4}$$ and $$8$$

$$4x^2-29x-24=0$$

##### 12.

$$\dfrac{5}{3}$$ and $$\dfrac{5}{3}$$

$$9x^2-30x+25=0$$

#### Exercise Group.

For Problems 13 and 14, graph the equation using the ZDecimal setting. Locate the $$x$$-intercepts, and use them to write the quadratic expression in factored form.

##### 13.

$$y=x^2-0.6x-7.2$$

$$y=(x-3)(x+2.4)$$

##### 14.

$$y=-x^2+0.7x+2.6$$

$$y=-(x+1.3)(x-2)$$

#### Exercise Group.

For Problems 15–18,

1. Find the coordinates of the vertex and the intercepts.

2. Sketch the graph.

##### 15.

$$y=\dfrac{1}{2}x^2$$

1. Vertex: $$(0,0)\text{,}$$ $$y$$-int: $$(0,0)\text{,}$$ $$x$$-int: $$(0,0)$$

2. ##### 16.

$$y=x^2-4$$

##### 17.

$$y=x^2-9x$$

1. Vertex: $$\left(\dfrac{9}{2},\dfrac{-81}{4}\right)\text{,}$$ $$y$$-int: $$(0,0)\text{,}$$ $$x$$-int: $$(0,0),~(9,0)$$

2. ##### 18.

$$y=-2x^2-4x$$

#### Exercise Group.

For Problems 19–22, solve by completing the square.

##### 19.

$$x^2-4x-6=0$$

$$2 \pm \sqrt{10}$$

##### 20.

$$x^2+3x=3$$

$$\dfrac{-3}{2} \pm \sqrt{\dfrac{21}{4}}$$

##### 21.

$$2x^2+3=6x$$

$$\dfrac{3}{2} \pm \sqrt{\dfrac{3}{4}}$$

##### 22.

$$3x^2=2x+3$$

$$\dfrac{1}{3} \pm \sqrt{\dfrac{10}{9}}$$

#### Exercise Group.

For Problems 23–26, solve the formula for the indicated variable.

##### 23.

$$K=\dfrac{1}{2}mv^2,~~$$for $$v$$

$$v=\pm \sqrt{\dfrac{2K}{m}}$$

##### 24.

$$a^2+b^2=c^2,~~$$for $$b$$

$$b=\pm \sqrt{c^2-a^2}$$

##### 25.

$$V=\dfrac{1}{3}s^2h,~~$$ for $$h$$

$$s=\pm \sqrt{\dfrac{3V}{h}}$$

##### 26.

$$A=P(1+r)^2,~~$$ for $$r$$

$$r=\pm \sqrt{\dfrac{A}{P}}-1$$

#### 27.

In a tennis tournament among $$n$$ competitors, $$\dfrac{n(n-1)}{2}$$ matches must be played. If the organizers can schedule 36 matches, how many players should they invite?

#### 28.

The formula $$S=\dfrac{n(n-1)}{2}$$ gives the sum of the first positive integers. How many consecutive integers must be added to make a sum of 91?

#### 29.

Lewis invested $2000 in an account that compounds interest annually. He made no deposits or withdrawals after that. Two years later he closed the account, withdrawing$2464.20. What interest rate did Lewis earn?

11%

#### 30.

Earl borrowed $5500 from his uncle for 2 years with interest compounded annually. At the end of two years he owed his uncle$6474.74. What was the interest rate on the loan?

8.5%

#### 31.

The perimeter of an equilateral triangle is 36 inches. Find its altitude. (Hint: The altitude is the perpendicular bisector of the base.)

$$\sqrt{108} \approx 10.4$$ in

#### 32.

The base of an isosceles triangle is one inch shorter than the equal sides, and the altitude of the triangle is two inches shorter than the equal sides. What is the length of the equal sides?

17 in

#### 33.

A car traveling at 50 feet per second (about 34 miles per hour) can stop in 2.5 seconds after applying the brakes hard. The distance the car travels, in feet, $$t$$ seconds after applying the brakes is $$d=50t-10t^2\text{.}$$ How long does it take the car to travel 40 feet?

1 sec

#### 34.

You have 300 feet of wire fence to mark off a rectangluar Christmas tree lot with a center-divider, using a brick wall as one side of the lot. If you would like to enclose a total area of 7500 square feet, what should be the dimensions of the lot?

50 ft by 150 ft

#### Exercise Group.

For Problems 35 and 36, show that the shaded areas are equal.

##### 35. $$A_1=x^2-\left(\dfrac{1}{2}y^2+\dfrac{1}{2}y^2\right)=x^2-y^2;~~A_2=(x+y)(x-y)=x^2-y^2$$ $$A_1=\pi (x+y)^2- \pi x^2 - \pi y^2 = 2 \pi xy;~~A_2=\pi y (2x) = 2 \pi xy$$