MAT 115 Course Prep

This collection of problems is meant to provide practice for a range of skills needed for MAT 115. Read and attempt each problem first; if you aren’t sure how to start a problem, explore the resources on the right to refresh your memory and try again. A drop-down button is found beneath each problem for you to compare both your logic and final answers. Keep track of the skills you aren’t comfortable with, and reacquaint yourself with them so you’re fully prepared for the topics you’ll grapple with soon. Get help in the Math Lab.  Return to the Course Prep page.

 

Problem 1: Using the functions defined below, simplify each expression. If possible, find exact values.

\boldsymbol{f(x)=x^2} \qquad \boldsymbol{g(x) = -x^2} \qquad \boldsymbol{h(x)=1-2x}
Part a. f(-1)

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f(-1) = (-1)^2 = 1.

Video: Worked Example: Evaluating Functions from Equation
Part b. g(-1)

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g(-1) = -(-1)^2 = -(1) = -1.

Part c. h(-1)

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h(-1) = 1-2(-1) = 1+2 = 3.

Part d. f(x)-g(x)

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f(x)-g(x) = (x^2) - (-x^2) = x^2+x^2=2x^2. Since no value of x is given, an exact value is not possible as an answer.

Part e. g(x)+f(x)

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g(x)+f(x) = (-x^2)+x^2 = 0.

Part f. \frac{f(x)}{g(x)}

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\frac{f(x)}{g(x)} = \frac{x^2}{-x^2} = -1, since the top and bottom share a common factor of x^2 which cancels (assuming x \neq 0).

Part g. f(x)g(x)

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f(x)g(x)=(x^2)(-x^2)=-x^4.

Part h. g(x)h(x)

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g(x)h(x) = (-x^2)(1-2x) = -x^2+2x^3, by the distributive property.

 

Problem 2: Solve the following equations for x, and verify your solution.

Part a. 100 = 0.2x

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Divide both sides by 0.2: 500=x. To verify, plug this value of x back into the original question and make sure that the statement is true. In this case, the choice of x = 500 gives us 100 = 0.2(500) \Rightarrow 100 = 100. Since this is true, we know our solution is correct.

To give you an idea of what it would look like if we somehow got the incorrect answer, pretend we solved the original equation and got x = 80. Then, in the verification step, we would have 100 = 0.2(80) \Rightarrow 100 = 16. This is obviously false, and so x= 80 is not a valid solution to the equation.

Video: Solving Linear Equations I

Write-up: One Step Equations

Part b. -80=\frac{4}{3}x

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Divide both sides by \frac{4}{3}. This has the effect of multiplying each side by \frac{3}{4}: \frac{-240}{4} = x, which reduces to show that x=-60. We verify by plugging this value back into the original question: -80 = \frac{4}{3}(-60) \Rightarrow -80 = \frac{-240}{3} \Rightarrow -80 = -80. This is true, and so we’ve verified our answer.

Part c. 44 = 8-6x

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Begin by subtracting 8 from each side, yielding 36 = -6x. Division by -6 gives x=-6. Verify the solution by plugging this value of x back into the original equation: 44 = 8-6(-6) \Rightarrow 44 = 8-(-36) \Rightarrow 44 = 8+36 \Rightarrow 44=44. This is true, and so we’ve verified our answer.

Video: Solving Linear Equations II

Write-up: Two Step Equations

Part d. \frac{x^2-6x}{x}=3

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On the left-hand side, a factor of x is common to each term in the numerator, and cancels with the x found in the denominator assuming x \neq 0: \frac{x^2-6x}{x} = \frac{x(x-6)}{x} = x-6. Simplification gives x-6=3, and adding 6 to both sides gives x=9. Verify the solution by substitution: \frac{(9)^2-6(9)}{9} = 3 \Rightarrow \frac{81-54}{9} = 3 \Rightarrow \frac{27}{9} = 3 \Rightarrow 3 = 3. This is true, and so we’ve verified our answer.

Part e. x^5=243

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To isolate the x, raise both sides to the \frac{1}{5} power. This gives x=243^{\frac{1}{5}} = \sqrt[5]{243} = 3. We plug in this value of x to verify the answer: (3)^5 = 243 \Rightarrow 3\cdot3\cdot3\cdot3\cdot3=243 \Rightarrow (9)\cdot3\cdot3\cdot3 = 243 \Rightarrow (27)\cdot3\cdot3 = 243 \Rightarrow \dots \Rightarrow 243 = 243. This is true, and so we’ve verified our answer.

Video: Solving Equations with Rational Exponents
Part f. x^9=27.8

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In a similar fashion as the previous problem, raise both sides to the \frac{1}{9} power, giving x=27.8^{\frac{1}{9}} = \sqrt[9]{27.8}. Using a calculator, we find that this is approximately 1.447.  To verify our answer, we can compute:  (1.447)^9 \approx 27.811 \approx 27.8 .  This shows that x = 1.447 is a correct approximation of the exact solution, which is  x = \sqrt[9]{27.8} .

Part g. xy=k

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In earlier questions, x was the only variable present, and “solving the equation” ended with a numerical solution. In cases like these, we must recall that “solving for x” is synonymous with isolating x on one side of the equation. In this equation, divide both sides by y: \frac{xy}{y} = \frac{k}{y} \Rightarrow x = \frac{k}{y} due to the cancellation on the left-hand side of the equation. We can still verify this solution via substitution for x in the original equation: \frac{k}{y} y = k \Rightarrow \frac{ky}{y} = k \Rightarrow k = k where the final step is possible due to the cancellation of y‘s on the left hand side. This statement is true, and so we’ve verified our answer.

Part h. y=xk

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In a similar fashion as the previous problem, we simply aim to isolate x on one side of the equation. Divide by k on both sides: \frac{y}{k} = \frac{xk}{k} \Rightarrow \frac{y}{k} = x. Verify via substitution: y = \frac{y}{k}k \Rightarrow y = \frac{yk}{k} \Rightarrow y = y, where the final step is possible due to the cancellation of k‘s on the right hand side. This statement is true, and so we’ve verified our answer.

 

Problem 3: Expand the following expressions.

Part a. (a+b)^2

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This can be rewritten as (a+b)(a+b), which can be FOILed (recall this tells you the order in which to multiply terms – First, Outer, Inner, Last). This gives (a)(a)+(a)(b)+(b)(a)+(b)(b) = a^2+2ab+b^2.

Video: Multiplying Binomials

Write-up: FOIL Method

Part b. (a-b)^2

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Following the same logic as the last problem, the multiplication gives (a)(a)+(a)(-b)+(-b)(a)+(-b)(-b) = a^2-2ab+b^2.

 

Problem 4: Find the slope of the line passing through the following points.

Part a. (0,5) and (4,19)

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Slope (commonly denoted by the symbol m) can be found via the familiar phrase “rise over run”. More accurately, we are concerned with the change in rise (y_2-y_1) over the change in run (x_2-x_1). In this case, this comes out to m = \frac{19-5}{4-0} = \frac{14}{4} = \frac{7}{2}.

Worked Example: Slope from Two Points

Write-up: Slope Formula

Part b. (2,11) and (3,-3)

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Following the same logic as the previous problem, we have m = \frac{-3-11}{3-2} = \frac{-14}{1} = -14.

Part c. (x_1,y_1) and (x_2,y_2)

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This represents the general case: m = \frac{y_2-y_1}{x_2-x_1}.

 

Problem 5: Find the equation of the line passing through the following points.

Part a. (0,5) and (4,19)

Click here to show solution.

There are two lanes of thought here, corresponding to the two different generic forms of a line: slope-intercept form [y = mx+b] and point-slope form [y-y_i = m(x-x_i)].

In slope-intercept form, the slope and y-intercept of the line need to be known (unsurprisingly). We calculated the slope between these points above: m = \frac{7}{2}. Separately, when x=0, we know that y=5 since (0,5) is given as a point on this line. Therefore, the y-intercept, commonly denoted with the symbol b, is 5. The equation of the line is therefore y=\frac{7}{2}x+5.

In point-slope form, the slope and any one point on the line must be known (unsurprisingly). Choose (x_i,y_i) = (0,5), for example: then y-5=\frac{7}{2}(x-0), which reduces to y-5=\frac{7}{2}x. Adding 5 to both sides gives the same answer as before: y=\frac{7}{2}x+5.

Point-slope form is generally more helpful, as needing any point on the line is a less strict requirement than knowing the y-intercept. The slope is needed in either case.

Video: Point-Slope and Slope-Intercept Equations

Write-up: Slope-Intercept Form

Write-up: Point-Slope Form

Part b. (2,11) and (3,-3)

Click here to show solution.

We don’t know the y-intercept here, so use point-slope form with either of the points provided. For the purpose of this example, we’ll use (2,11); recall from the earlier question that the slope between these points is -14. The equation of the line is y-11=-14(x-2) = -14x+28. Adding 11 to both sides gives y=-14x+39. (Note that from this manipulation, we can actually determine the y-intercept to be 39, as the final simplified answer is in slope-intercept form with m=-14, b=39. This isn’t obvious from the information given initially.)

Part c. (x_1,y_1) and (x_2,y_2)

Click here to show solution.

Use point-slope form again. For slope, recall that m = \frac{y_2-y_1}{x_2-x_1}, and note that either point can be plugged into the point-slope formula. Both y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) and y-y_2=\frac{y_2-y_1}{x_2-x_1}(x-x_2) are correct.

 

Problem 6: Your annual salary is $40,000 for the year 2018. Use the information below to calculate your salary in 2019, 2020, and 2021.

Part a. Assume your salary increases by $1500 per year.

Year 2018 2019 2020 2021
Salary $40,000 ??? ??? ???

 

Click here to show solution.

Take the salary for each year and add 1500 to get the salary for the next year.

Year 2018 2019 2020 2021
Salary $40,000 $41,500 $43,000 $44,500
Part b. Assume your salary increases by 3% per year.

Year 2018 2019 2020 2021
Salary $40,000 ??? ??? ???

 

Click here to show solution.

Let’s start with the calculation for 2019. An increase of 3% over the salary in 2018 means that 40,000+\frac{3}{100}(40,000) is earned in 2019. Factor 40,000 out of the left hand side to find 40,000(1+\frac{3}{100})=40,000(1+0.03) = (1.03)(40,000) = 41,200. This logic can be applied recursively for the other future years:

Year 2018 2019 2020 2021
Salary $40,000 (1.03)(40,000) = $41,200 (1.03)(41200) = $42,436 (1.03)(42436) = $43,709.08

 

Problem 7:  Use the following graph to answer the questions below.

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MathPlanet – Coordinates & Ordered Pairs

Part a. Does the point (6, 5) fall on the graph of the line ?

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A coordinate point tells you the x-coordinate first, followed by the y-coordinate.  We can see that the point (6, 5) has an x-coordinate of 6.  We can see on the x-axis, between 0 and 5, that this region is divided into 5 equal parts, marked by the light grey lines.  This means that each square cell on the grid is one unit wide.  To find 6 on the x-axis, we just need to go one unit to the right of the 5.  Then, we follow the light grey vertical grid-line up to the red line, and we have found the point on the line with x-coordinate 6.  Finally, we follow the light grey horizontal grid-line from this point to the left, until we reach the y-axis.  This horizontal grid-line intersects the y-axis right at the number 5, so we know that 5 is the y-coordinate that corresponds to the x-coordinate of 6.

Thus, we have determined that (6, 5) does in fact fall on the graph of the line.

 

Problem 8:  Rewrite each expression in the form:  xa   (this is called base-exponent form).

Part a. \dfrac{1}{x^{2}}

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We can move a factor from the denominator to the numerator as long as we change the sign of the exponent:

\dfrac{1}{x^{2}} = \dfrac{x^{-2}}{1}

If it seems like the 1 in the numerator disappeared, remember that multiplying by one won’t change the value of x^{-2}  .  Dividing by 1 also won’t change the value of x^{-2}  , so we can just write:

\dfrac{1}{x^{2}} = \boxed{x^{-2}}

If we want to show all the steps, we could write:

\dfrac{1}{x^{2}} = \dfrac{1}{1 \cdot x^{2}} = \dfrac{1 \cdot x^{-2}}{1} =\dfrac{x^{-2}}{1} = x^{-2}

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BrownMath – Negative and Fractional Exponents

MathPlanet – Exponent Rules

Part b. \sqrt{x}

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We can rewrite radical expressions using fractional exponents.  For a square root, we use the fraction 1/2:

\sqrt{x} = \boxed{x^{\frac{1}{2}}}

If this seems strange, think of it this way:  20 = 1  and  21 = 2  so we would expect that  2^{\frac{1}{2}}   would produce a number between 1 and 2, and  \sqrt{2}   is between 1 and 2.

Part c. \left(\dfrac{1}{x^{3}}\right)^{4}

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We can distribute an exponent over division, which means the 4 becomes the exponent of the 1 in the numerator and the x^3 in the denominator:

\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{(1)^{4}}{\left(x^{3}\right)^{4}}

Now 1 multiplied by itself 4 times is still 1, but in the denominator we have to combine the exponents.  Recall that successive exponents can be combined by multiplication:

\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{12}}

Finally, we can rewrite this by moving the x into the numerator, and changing its exponent from positive to negative:

\left(\dfrac{1}{x^{3}}\right)^{4} = \boxed{x^{-12}}

We could also have written out 4 copies of the original fraction:

\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}} \cdot \dfrac{1}{x^{3}}

This could then be expanded as:

\left(\dfrac{1}{x^{3}}\right)^{4} = \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x} \cdot \dfrac{1}{x\cdot x\cdot x}

Part d. \sqrt[3]{x^{2}}

Click here to show solution.

Recall that the 3 is called the index of the root, and indicates that this is not a square root but a cube root.  Following our earlier rule, we can rewrite this root using a 1/3 exponent:

\sqrt[3]{x^{2}} = \left(x^{2}\right)^{\frac{1}{3}}

Exponent rules allow us to multiply successive exponents, so we calculate 2 times 1/3 to get 2/3 as the exponent:

\sqrt[3]{x^{2}} = \boxed{x^{\frac{2}{3}}}.

 

Problem 9: Expand the following expressions. Remember to use FOIL.

F.O.I.L. stands for First, Outer, Inner, Last.

When two binomials are being multiplied together, we can use FOIL to remember all the products that appear in the expansion.

For example:   (a + b)(c + d)  =  ac + ad + bc + bd

\boxed{ac}    is the product that results from multiplying the First terms together.

\boxed{ad}    is the product of the Outer terms.

\boxed{bc}    is the product of the Inner terms.

\boxed{bd}    is the product of the Last terms.

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MathPlanet – Multiplication of Polynomials

Part a. (x+4)(x+6)

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Following FOIL will give us the following:

x \cdot x + x \cdot 6 + 4 \cdot x + 4 \cdot 6

We can take each product and simplify it as:

x^{2} + 6x + 4x + 24

Now we can combine 6x and 4x to make 10x:

\boxed{x^{2} + 10x + 24}.

Part b. (x+8)(x-3)

Click here to show solution.

Following FOIL will give us the following:

x \cdot x + x \cdot -3 + 8 \cdot x + 8 \cdot -3

We can take each product and simplify it as:

x^{2} - 3x + 8x - 24

Now we can combine -3x and 8x to make 5x:

\boxed{x^{2} + 5x - 24}.

Part c. (-x+2)(x-7)

Click here to show solution.

Following FOIL will give us the following:

-x \cdot x + -x \cdot -7 + 2 \cdot x + 2 \cdot -7

We can take each product and simplify it as:

-x^{2} + 7x + 2x - 14

Notice how -x times -7 gave us a positive result, because the negatives cancel each other.

Now we can combine 7x and 2x to make 9x:

\boxed{x^{2} + 9x - 14}.

Part d. (-x-1)(-x+5)

Click here to show solution.

Following FOIL will give us the following:

-x \cdot -x + -x \cdot 5 + -1 \cdot -x + -1 \cdot 5

We can take each product and simplify it as:

x^{2} - 5x + 1x - 5

Notice how the first term is positive, because it is the product of two negatives.  This is also true of the third term, 1x.  Also, instead of writing 1x, we usually just write x, and the coefficient of 1 is implied

Now we can combine -5x and x to make -4x:

\boxed{x^{2} - 4x - 5}.

 

Problem 10:  Rewrite each expression in factored form.

Part a. x^{2}+8x+12

Click here to show solution.

List the factors of the constant term, 12:

\begin{array}{c|c} &  \\\hline 1 & 12  \\2 & 6  \\3 & 4  \\-1 & -12  \\-2 & -6  \\-3 & -4  \end{array}

Notice that we include pairs where both numbers are negative.

We want to choose the pair that adds up to the coefficient of x, which is 8, so we pick 2 and 6.

x^{2}+8x+12 = \boxed{(x+2)(x+6)}.

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MathPlanet – Factoring Quadratics

Part b. x^{2}+9x+8

Click here to show solution.

List the factors of the constant term, 8:

\begin{array}{c|c} &  \\\hline 1 & 8  \\2 & 4  \\-1 & -8  \\-2 & -4  \end{array}

Notice that we include pairs where both numbers are negative.

We want to choose the pair that adds up to the coefficient of x, which is 9, so we pick 1 and 8.

x^{2}+9x+8 = \boxed{(x+1)(x+8)}.

Part c. x^{2}+2x-15

Click here to show solution.

List the factors of the constant term, -15:

\begin{array}{c|c} &  \\\hline 1 & -15  \\3 & -5  \\-1 & 15  \\-3 & 5  \end{array}

Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative.

We want to choose the pair that adds up to the coefficient of x, which is 2, so we pick -3 and 5, because adding them together gives us -3 + 5 = 2.

x^{2}+2x-15 = \boxed{(x-3)(x+5)}.

Part d. x^{2}-x-42

Click here to show solution.

List the factors of the constant term, -42:

\begin{array}{c|c} &  \\\hline 1 & -42  \\2 & -21  \\3 & -14  \\6 & -7  \\-1 & 42  \\-2 & 21  \\-3 & 14  \\-6 & 7  \end{array}

Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative.

We want to choose the pair that adds up to the coefficient of x, which is -1, so we pick 6 and -7, because when we add them together we get 6 + -7 = -1.

x^{2}-x-42 = \boxed{(x+6)(x-7)}.

Part e. x^{2}-7x+12

Click here to show solution.

List the factors of the constant term, 12:

\begin{array}{c|c} &  \\\hline 1 & 12  \\2 & 6  \\3 & 4  \\-1 & -12  \\-2 & -6  \\-3 & -4  \end{array}

Notice that we include pairs where both numbers are negative.

We want to choose the pair that adds up to the coefficient of x, which is -7, so we need to pick two negative numbers.  We pick -3 and -4 because adding them gives us -3 + -4 = -7

x^{2}-7x+12 = \boxed{(x-3)(x-4)}.

Part f. x^{2}+3x-40

Click here to show solution.

List the factors of the constant term, -40:

\begin{array}{c|c} &  \\\hline 1 & -40  \\2 & -20  \\4 & -10  \\5 & -8  \\-1 & 40  \\-2 & 20  \\-4 & 10  \\-5 & 8  \end{array}

Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative.

We want to choose the pair that adds up to the coefficient of x, which is 3, so we pick -5 and 8, because when we add them together we get -5 + 8 = 3.

x^{2}+3x-40 = \boxed{(x-5)(x+8)}.

 

Problem 11:  Solve the linear inequality.

Part a.  4x + 5 < 7x - 13

 

Click here to show solution.

Notice that there are terms containing x on both sides of the inequality.  First, we can subtract 4x from both sides so that we can combine these like terms:

5 < 3x - 13

Now we can move all the constants to the other side, which means we must add 13 to both sides:

18 < 3x

Finally, we can divide both sides by 3 to isolate the variable:

6 < x

This says 6 is less than x, which is another way of saying that x is greater than 6.

Thus, we have:  \boxed{x > 6}. 

Need Help ?

Video: KhanAcademy – Solving Inequalities

 

Problem 12:  Express the following using interval notation.

Part a. 7 < x < 13

 

Click here to show solution.

The statement literally says that 7 is less than x, which is less than 13.

In other words, x is greater than 7 and x is less than 13.

This indicates a single interval, from 7 to 13.

Because the original statement uses \boxed{<}  and not \boxed{\leq}  we know that 7 and 13 are NOT included in the interval.

We use “soft brackets” (parentheses) to indicate that the interval includes all numbers between 7 and 13, but does NOT include the endpoints:

\boxed{( 7, 13 )}.

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Video: KhanAcademy – Interval Notation

Text: Lumen – Interval Notation

Part b. -16 \leq x \leq 10

 

Click here to show solution.

The statement literally says that -16 is less than or equal to x, which is less than or equal to 10.

In other words, x not greater than 10, and is not less than -16.

This indicates a single interval with endpoints at -16 and 10.

Because the original statement uses \boxed{\leq}  and not \boxed{<}  we know that -16 and 10 ARE included in the interval.

We use “hard brackets” [square braces] to indicate that the interval includes all numbers between 7 and 13 and ALSO includes the endpoints:

\boxed{[ -16, 10 ]}.

Part c. -8 \leq x < -4

 

Click here to show solution.

Compared to the previous two problems, we simply need to use a hard bracket to start our interval, because the -8 is NOT included, and a soft bracket to end the interval, because the -4 IS included.

\boxed{[ -8, -4 )}.

Part d. x > 56

 

Click here to show solution.

The statement says x is greater than 56.  This excludes the possibility that x = 56, so we will use a soft bracket.

All the numbers greater than 56 together form a sort of interval with a left endpoint at 56, but no right endpoint.  Since the interval starts at 56 and never ends, we can think of the right endpoint as being “at” infinity.

Infinity is not a number, so we do not want to include it, which means we must use a soft bracket here as well.  Using infinity as our right endpoint means that all numbers between 56 and infinity are included, even if both endpoints are not.

\boxed{( 56, \infty )}.

Part e. x \leq -42

 

Click here to show solution.

The statement says that x is less than or equal to -42.  This includes the possibility that x = -42, so we must use a hard bracket.

Since all negative numbers less than -42 are included in this interval, we need to start our interval at negative infinity (but not include it).  -42 is actually where our interval ends.

In other words, negative infinity is the left endpoint of this interval, and -42 is the right endpoint.

\boxed{( -\infty, -42 ]}.

 

Problem 13:  Solve using the quadratic formula to get exact answers (expressed using radicals).

Part a. x^2 - 8x + 5 = 0

 

Click here to show solution.

Recall the quadratic formula:

x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}

In the quadratic equation given above, we can see that a = 1,  b = -8,  and c = 5.

Notice that, even thought there is no visible coefficient in front of the x2 term, we always assume a coefficient of 1, which is why a = 1.  Also, the minus sign in front of the 8 must be included in the b value, which is why we have b = -8.

x = \dfrac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(5)}}{2(1)}

Notice how we placed parentheses around every place where we plugged in a value.  Now we want to simplify  -(-8) = 8  and  (-8)2 = 64  and  4(1)(5) = 20  and  2(1) = 2.

x = \dfrac{8 \pm \sqrt{64-20}}{2}

Since  64 – 20 = 44,  and  44 = 4 times 11,  we can write:

x = \dfrac{8 \pm \sqrt{4 \cdot 11}}{2}

4 is a square number, so we can pull it out from under the square root as a 2:

x = \dfrac{8 \pm 2\sqrt{11}}{2}

Now, we can divide by 2.  Recall that each term in the numerator must be divided by 2:

x = 4 \pm \sqrt{11}

Finally, recall that the symbol  \boxed{\pm}    means that we actually have two separate solutions for x:

\boxed{4 + 2\sqrt{11} \quad \text{or} \quad 4 - 2\sqrt{11}}.

Need Help ?

Video: KhanAcademy – Quadratic Formula

 


 

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