Quick algebra review by peter selby

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1. In the standard form of a quadratic equation, ax 2 + bx + c = 0, a represents the numerical coefficient of x 2 (the second degree term of the unknown);b is the numerical coefficient of x (the first degree term of the unknown);and c represents the constant term. In other words, a, 6, and c represent the real number coefficients, and x represents the variable. In the equation x 2 - 3x + 2 = 0, a = 1, b = - 3, and c = 2

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2. Quadratic equations are divided into two classes: complete and incomplete quadratic equations. A complete quadratic equation contains all three terms;an incomplete quadratic equation has one term missing. For example, in the equation x 2 - 16 = 0, the x term is missing in the equation; x 2 -6x =0, the constant term is missing.Indicate which of the following are complete and which are incomplete.(a) x 2 + x - 2 = 0 (c) 3x 2 = 2x (b) 2x-l = x 2 (d) Ax 2 - 49 = 0(a) complete; (b) complete (but needs rearranging); (c) incomplete (lacks constant); (d) incomplete (lacks x term)

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3. Although there are procedures to assist you in factoring a trinomial (as discussed in Unit 4), factoring some types of trinomials is a matter of educated guessing.Consider the trinomial x 2 + 2x - 15. Since the numerical coefficient of x 2 is 1, the first term in each of the two binomial factors will simply be x. The next step is to examine the factors of 15 to discover if any two of them differ by 2 (the numerical coefficient of the middle term of the trinomial). Since 3 times 5 and 1 times 15 are the only integral factors of 15, it is evident that the correct pair of factors is 3 and 5

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4. Having reviewed the procedure for factoring quadratic expressions, we can consider how to solve equations involving quadratic expressions. There are several methods available. The first one we will review is that of solution by factoring. The method is outlined in review item 4. Here is another example.Example: Solve: x 2 - 6 = xSolution: Rewrite in standard form: x 2 - x - 6 = 0 Factor left member: (x + 2)(x - 3) = 0 Set each factor equal to zero: x+2 = 0;x-3 = 0 Solve the resulting firstdegree equations: x = -2; x = 3 (hence -2 and 3 are the roots)Check: (-2) 2 - 6 = -2; 4 - 6 = -2 -2 = -2

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(3)2 - 6 = 3; 9-6 = 3 3 = 3Note that the third step of this procedure uses the concept called the zero factor law. This law states that if a • b = 0, then a = 0, b = 0, or both a and b = 0. This concept applies only when one member of an equation is zero.Solve the following quadratic equations by the factoring method. Check each of the roots in the original equation from which it was derived. You will not know for certain whether or not you have made a mistake in deriving a numerical value as a root unless you test it in the original equation.(a) x 2 + 6x + 8 = 0 (e) 3x 2 = x (b) 4x 2 - 7x - 2 = 0 (f) 2y 2 - 5y = 25(c) x 2 - 9 = 0 (g) k 2 - Ak = 0 (d) 9x 2 - 6x = -1

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5. It is apparent that the method of solution by factoring will not always work, since not all quadratic equations are factorable. Now we review the procedure for solving incomplete quadratic equations in which the constant term equals zero. As indicated in review item 2, such equations take the form ox 2 + bx = 0.If we factor the left member of such an equation, we get x(ax -I- 6) = 0, from which x = 0 and ox + b = 0, representing two linear equations. If we solve the second equation for x, we get x = - —; hence, the roots of such an equation b^ b aare zero and - —.aExample: Find the roots of the equation y 2 - 2y = 0.Solution: We can solve this equation by factoring, which gives usy(y - 2) = 0, from which y = 0, and y - 2 = 0, y = 2. On the other hand, from our general solution above, we can simply write at once the second solution asy = - — = 2 (since in this case a — 1 and b = -2). -2Solve the following equations using either the factoring method or the method based on the general solution.(a) 4x 2 = 28x (remember to change to standard form ox 2 + bx = 0)

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6. Solving an incomplete quadratic equation of the type ax 2 = c where the coefficient of the first degree term is zero, requires a somewhat different ap- proach, since factoring may or may not work. If factoring does not work, we use the method of extraction of roots as summarized below:'3 b— 5 ô*•-;>Step 1: Solve for the square of the variable. This will yield an equation of the form x 2 = —.ac Ic Step 2: The roots of x 2 = — will be the roots of the two equations x =*/— andIF c ^x if ~ is positive. (There will be no real number solution if^ is negative.)Step 3: Check results back in the original equation.Example: Solve the equation 9x 2 - 25 = 0 using the method of extraction of roots.Solution: Step 1: 9x 2 = 25, x 2 = — 25„ „ [25 [25 u 5 5 5Step 2: x =W — and x = - y —, hence x = - and - - or ± - Step 3: Check: 9 ( - j 2 25 = 0; 9 ( y I - 25 = 0; 0 = 0.Solve the following equations by the method of extraction of roots. In prob- lems (c) and (d), start by dividing both terms by the coefficient of x 2

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7. A quadratic equation is not always in the standard form ox 2 + bx + c = 0. In fact, sometimes you will have to change the form of an equation just to see whether it is a quadratic equation. For example, x = 4 - — may not at 3 first appear to be a quadratic equation. However, clearing fractions and transforming the resulting equation gives us x 2 - 4x + 3 = 0, which is somewhat more recognizable.If a quadratic equation is not in standard form, perform whatever operations are necessary to transform it to standard form. Any of the axioms we have developed for transforming to equivalent equations, such as those listed in review item 7, can be used.Use any procedures necessary to express the following quadratic equations in standard form. Write the equation with a positive coefficient for the second degree term.(a) Jy 2 - 5y = 3y (b) 20 + 6* = 2k 2+ 1 = 4c (d) p 2 — 5p — 4(e) 3b 2 = -5b(f) 7(x 2 - 9) = x(x - 5) (g) 18 = 2x 2(h) y(8 - 2y) = 6(a)8y 2 + 5y = 0; (b)2k 2 -6k-20 = 0; (c)4c 2 -c-10 = 0; (d)p 2 -5p + 4 = 0;(e) 36 2 + 56 = 0; (f)6x 2 + 5x - 63 = 0; (g)2x 2 -18 = 0; (h) 2y 2 - 8y + 6 = 0

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8. We have mentioned that some quadratic equations cannot be solved by factoring. One method that can be used to solve such equations is called com- pleting the square. This method requires that you make one member of the equation a perfect square. For example, the left member of the equation x 2 + 6x - 7 = 0 is not a perfect square trinomial because the third term, -7, does not have the correct value. However, the binomial x 2 + 6x could be converted to a perfect square trinomial by adding a third term equal to the square of one-half the coefficient of x, that is, by adding 3 2 or 9. The necessary steps for doing this and working out the final solution of the equation are shown in review item 8

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11. The last method we will discuss for solving quadratic equations is that of graphing.In Unit 8, we reviewed the procedure for plotting a curve, as all plots of equations are termed, regardless of whether they are straight lines or curved.We found that, in plotting a curve on a rectangular coordinate system, we needed a pair of coordinates to locate any single point on the curve. These coordinates are the x coordinate (abscissa) and y coordinate (ordinate). The same general procedure for plotting a curve outlined in Unit 8 can be used to

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12. The graph of a quadratic equation does not always cross the X axis. Sometimes it is tangent to it (that is, it touches the X axis at only one point).In that case, it will have two equal, real roots, more correctly expressed by saying that it will have one root with a multiplicity of 2. Observe this in the example below. The curve shown is tangent to the X axis. The roots are x = 2, 2

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Tiêu đề	Quick Algebra Review
Tác giả	Peter H. Selby
Người hướng dẫn	Judy V. Wilson, Alicia Conklin, Maria Colligan
Trường học	John Wiley & Sons, Inc.
Chuyên ngành	Algebra
Thể loại	self-teaching guide
Năm xuất bản	1983
Thành phố	San Diego

Định dạng
Số trang	241
Dung lượng	8,75 MB