MATH: COLLEGE ALGEBRASection III

DePaul Math Placement Test

Absolute valuesTo solve an equation in which the variable is within absolute value brackets, you must divide the equation into two equations. The two equations are necessary because an absolute value really defines two equal values, one positive and one negative. The most basic example of this is an equation of the form |x| = c. which is equivalent to x = c or-c.

A slightly more complicated example is this:|x + 3| = 5. Solve for x.In this problem, you must solve two equations: First, solve for x in the equation x + 3 = 5. In this case, x = 2. Second, solve for x in the equation x + 3 = –5. In this case, x = –8. So the solutions to the equation |x + 3| = 5 are x = {–8, 2}.To solve an equation in which the variable is within absolute value brackets, •isolate the expression within the absolute value brackets, and then create two equations. •Keep one of these two equations the same, while in the other equation negate one side of the equation. •In either case, the absolute value of the expression within brackets will be the same. •This is why there are always two solutions to absolute value problems (unless the variable is equal to 0, which is neither positive nor negative).

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Absolute values

Here is one more example:Q. Solve for x in terms of y in the equation |( x+2)/3|= (y2 – 1)/3

A. First, isolate the expression within the absolute value brackets:

|(x+2)/3|= (y2 – 1). Then solve for the variable as if the expression within absolute value brackets were positive: x+2 = y2 - 1 3 3 x+2 = y2 - 1 x = y2-3Next, solve for the variable as if the expression within absolute value brackets were negative: x+2 = - (y2 – 1) 3 3 x+2 =- y2 + 1 x+2 = y2 +1 x = -y2 -1The solution set for x is {y2 – 3, –y2 –1}.

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Inequalities

Before you get too comfortable with expressions and equations, we should

introduce inequalities. An inequality is like an equation, but instead of relating

equal quantities, it specifies exactly how two expressions that are not equal.x > y - “x is greater than y.”x < y - “x is less than y.”x ≥ y - “x is greater than or equal to y.”x ≤ y - “x is less than or equal to y.”Solving inequalities is exactly like solving equations except for of one verey

important difference:

when both sides of an inequality are multiplied or divided by a negative number, the relationship between the two sides changes and so the direction of the inequality must be switched.

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Inequalities

Here are a few examples: Solve for x in the inequality: (x/2) – 3 < 2y (x/2) < 2y + 3 x<2(2y+3) x<4y+6 (4/x)< -2 4≥ -2x 2 ≤ x Notice that in the last example, the inequality had to be reversed. Another way to express the solution is x ≥ –2. To help you remember that multiplication or division by a negative number reverses the direction of the inequality, recall that if x > y, then, just as 5 > 4 and –5 < –4. Intuitively, this makes sense, and it might help you remember this special rule of inequalities.

The number of solutions to an equation is usually equal to the highest power of the equation. A linear equation (highest term of x) has one solution, a quadratic equation (highest term of x2) has two solutions, and a cubic equation (highest term of x3) has three solutions. In an inequality, this rule does not hold true.

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Absolute Value and Inequalities

An equation without any absolute values generally results in, at most, only a few different solutions. Solutions to inequalities are often large regions of the x-y plane, such as x < 5. As we’ve seen before, inequalities produce usually introduces two sets of solutions. The same is true when absolute values are introduced to inequalities: the solutions often come in the form of two regions of the x-y plane.

If the absolute value is less than a given quantity, then the solution is a single range, with a lower and an upper bound. Q. Solve for x in the inequality |2x – 4| ≤ 6.

A. First, solve for the upper bound: 2x – 4 ≤ 6 2x≤10 x≤10 x≤5 Second, solve for the lower bound: 2x -4≥ -6 2x≥-2 x≥ -1Now combine the two bounds into a range of values for x {–1 ≤ x ≤ 5 }.

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Absolute values and Inequalities

Q. Solve for x in the inequality |3x + 4| > 16A. First, solve for the upper range:

3x+4>16 3x>12 x>4

Then, solve for the lower range:3x +4<-163x<-20 x<-20/3Now combine the two ranges to form the solution, 4<x<-20/3.

The other solution for an absolute value inequality involves disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and the second whose lower bound is a real number and whose upper bound is infinity. This occurs when the absolute value is greater than a given quantity which is two disjoint ranges: –∞ < x < – 20/3 or 4 < x < ∞.

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Ranges

Inequalities are also used to express the range of values that a variable can take. a < x < b means that the value of x is greater than a and less than b.

Consider the following word problem example:

Q “This game is appropriate for people older than 40, but no older than 65.” What is the range of the age of people for which the board game is appropriate?

A. Let a be the age of people for whom the board game is appropriate. The lower bound of a is 40, and the upper bound is 65.

The range of a does not include its lower bound (it is appropriate for people “older than 40”), but it does include its upper bound (“no older than 65,” i.e., 65 is appropriate, but 66 is not).

Therefore, the range of the age of people for which the board game is appropriate can be expressed by the inequality:

40≤ a≤65

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Ranges

Every time you come across a question involving ranges, you should carefully scrutinize the problem to pick out whether or not a particular variable’s range includes its bounds. This inclusion is the difference between “less than or equal to” and simply “less than.” Operations like addition, subtraction, and multiplication can be performed on ranges just as they are performed on variables or inequalities. For example:If 4 < x < 7, what is the range of 2x + 3?To solve this problem, simply manipulate the range like an inequality until you have a solution. Begin with the original range:4<x<7Then multiply the inequality by 2:

8<2x<14Add 3 to the inequality, and you have the answer.

11<2x+3<17There is one crucial rule that you need to know about multiplying ranges: if you multiply a range by a negative number, you must flip the greater than or less than signs. For instance, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > x > –8.

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Quadratic Equations

A quadratic, or quadratic polynomial, is a polynomial of the form ax2 + bx + c, where a ≠ 0. The following polynomials are quadratics: x2+3x =4, x2 -6x, t2+ 1, u2

A quadratic equation sets a quadratic polynomial equal to zero. That is, a quadratic equation is an equation of the form ax2 + bx + c = 0. The values of x for which the equation holds are called the roots, or solutions, of the quadratic equation. Most of the questions on quadratic equations involve finding their roots.There are two basic ways to find roots: by factoring and by using the quadratic formula. Factoring is faster, but it can’t always be done. The quadratic formula takes longer to work out, but it works for all quadratic equations. We’ll study both in detail.

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Factorization

To factor a quadratic, you must express it as the product of two binomials. In essence, factoring a quadratic involves a reverse-FOIL process. Take a look at this quadratic:

x2+10x+21 In the example above, the leading term has a coefficient of 1 (1x2 is the same as x2). •Since the two x variables are multiplied together during the FIRST step of foiling to get the first term of the quadratic polynomial, we know that the binomials whose product is this quadratic must be of the form (x+m)(x+n), where m and n are constants. •You also know that the sum of m and n is 10, since the 10x is derived from multiplying the OUTER and INNER terms of the binomials and then adding the resulting terms together (10x = mx + nx, so m + n must equal 10). Finally, you know that the product of m and n equals 21, since 21 is the product of the two last terms of the binomials.•Now to find the values of m and n. You know that x is the first term of both binomials, and you know that the sum of m and n is 10 and the product of m and n is 21. The pair of numbers that fit the bill for m and n are 3 and 7. Thus, x2 + 10x + 21 = (x + 3)(x + 7). f you have such an equation, then once you have factored the quadratic you can solve it. Thus, since x + 3 = 0 or x + 7 = 0, the solutions (also known as the roots) of the quadratic must be x = –3 and x = –7.

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Quadratics with Negative Terms

So far we’ve seen only with quadratics in which the terms are all positive. Factoring a quadratic that has negative terms is no more difficult, but it might take slightly longer to get the hang of it, simply because you are less used to thinking about negative numbers. Q. Consider the quadratic equation x2 – 4x – 21 = 0.A. There are a number of things you can tell from this equation: the first term of each binomial is x, since the first term of the quadratic is x2; the product of m and n is –21; and the sum of a and b equals –4. The equation also tells you that either m or n must be negative but that both cannot be negative, because the multiplication of one positive and one negative number can only result in a negative number. Now you need to look for the numbers that fit these requirements for m and n. The numbers that multiply together to give you –21 are: –21 and 1, –7 and 3, 3 and –7, and 21 and –1. The pair that works in the equation is –7 and 3.

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Polynomial Identities

There are two special quadratic polynomials that pop up quite frequently on the placement test and also in Calculus, and you should memorize them.

They are the perfect square and the difference of two squares. If you memorize the formulas below, you may be able to avoid the time taken by factoring.

There are two kinds of perfect square quadratics. They are:• a2 +2ab + b2 = (a + b)(a + b) = (a + b)2. Example: a2 + 6ab + 9 = (a + 3)2• a2 – 2ab + b2 = (a – b) (a – b) = (a – b)2. Example: a2 – 6ab + 9 = (a –3)2Note that when you solve for the roots of a perfect square quadratic equation, the solution for the equation (a + b)2 = 0 will be –b, while the solution for (a + b)2 = 0 will be b.

 

The difference of two squares quadratics follow the form below:

(a+b)(a-b) = a2-b2. Example: (a+3)(a-3) = a2-9 

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Polynomial Identities

Q. Here’s an instance where knowing the perfect square or difference of two square equations can help you: 2x2 + 20x + 50 = 0 A.To solve this problem by working out the math, you would do the following:2x2 + 20x + 50 = 0 2(x2 + 10x + 25) = 0 (x+5)2= 0 x = -5If you got to the step where you had 2(x2 + 10x +25) = 0 and realized that you were working with a perfect square of 2(x + 5)2, you could immediately have divided out the 2 from both sides of the equation and seen that the solution to the problem is –5.

Practice QuadraticsTake a look at the following examples and try to factor them on your own before you peek at the answers. x2+ x-2 = 0 Roots: {-2.1} x2 + 13x + 42 = 0 Roots: {-7,-6} x2 -8x+15= 0 Roots:{3,5} 

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The Quadratic Formula

Factoring using the reverse-FOIL method is really only practical when the roots are integers. Quadratics, however, can have decimal numbers or fractions as roots. Equations like these can be solved using the quadratic formula. For an equation of the form ax2 + bx + c = 0, the quadratic formula states:

 

Consider the quadratic equation x2 + 5x + 3 = 0. There are no integers with a sum of 5 and product of 3. So, this quadratic can’t be factored, and we must resort to the quadratic equation. We plug the values, a = 1, b = 5, and c = 3 into the formula:

 

 

,

The roots of the quadratic are approximately {–4.303, –.697}.

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Quadratic Equations- Finding Discriminant

If you want to find out quickly how many roots an equation has without calculating the entire formula, all you need to find is an equation’s discriminant. The discriminant of a quadratic is the quantity b2 – 4ac. As you can see, this is the radicand in the quadratic equation.

If:•b2 – 4ac = 0, the quadratic has one real root and is a perfect square.•b2 – 4ac > 0, the quadratic has two real roots.•b2 – 4ac < 0, the quadratic has no real roots, and two complex roots.

This information is useful when deciding whether to crank out the quadratic formula on an equation, and it can spare you some unnecessary computation. For example, say you’re trying to solve for the speed of a train in a rate problem, and you find that the discriminant is less than zero. This means that there are no real roots (a train can only travel at speeds that are real numbers), and there is no reason to carry out the quadratic formula.

Now, test your knowledge of the topics discussed by clicking on the sampletest links given below.

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