Welcome to Math 19500 Video Lessons Prof. Department of Mathematics The City College of New York Fall 013 An important feature of the following Beamer slide presentations is that you, the reader, move step-by-step and at your own pace through these notes. To do so, use the arrow keys or the mouse to move from slide to slide, forwards or backwards. Also use the index dots at the top of this slide (or the index at the left, accessible from the Adobe Acrobat Toolbar) to access the different sections of this document. To prepare for the Chapter 1.5 Quiz (September 5th at the start of class), please Read all the following material carefully, especially the included Examples. Memorize and understand all included Definitions and Procedures. Work out the Exercises section, which explains how to check your answers. Do the Quiz Review and check your answers by referring back to the Examples.
Linear and polynomial equations Definition An equation is a statement that two algebra expressions are equal. Numerical equations (involving numbers no variables) are either true or false. For example, 3 + 5 = 8 is true, while 3 + 5 = 0 is false. For an equation with a single variable x, there are three possibilities: true for all x, as is the case for the equation x + = + x, true for some but not all x, as is the case for the equation x + = 9 false for all x, as is the case for the equation x + = x + 1. Definition A real number is a solution of an equation in x if substituting that number for x in the equation yields a true statement. For example, 6 is a solution of x + 3 = 9 because (6) + 3 = 9 is a true statement. In fact, 6 is the only solution of x + 3 = 9.
However, equations can have more than one solution. Check that and are solutions of x = 4, and that 0, 1, and are solutions of x 3 + x = 3x. Definition A polynomial equation in x is a statement P (x) = Q(x) that two polynomials in x are equal. The degree of a polynomial equation P (x) = Q(x) is the highest power of x in the polynomial p(x) = P (x) Q(x) when p(x) is written in standard form. Example 1: The equation 1 + x + x + x 3 = has degree 3. The equation x + 3x = x 5 has degree one since it can be rewritten as 3x + 5 = 0. Definition A linear equation in x is a degree one polynomial equation in x.
A linear equation in x can always be written in the form p(x) = mx + b = 0, where m and b are real numbers, and the graph of the equation y = p(x) is a straight line with slope m. Examples of linear equations: 3x = 5 3x + 5 = 0 x + 3x = x + 5 Procedure To solve a linear equation in x: Rewrite the equation in the form ax = b. The solution is then x = b/a. Example : To solve 3x + 5 = 8x, rewrite the equation as 5x = 7. The solution is x = 7/5. Example 3: Solve xc + yc + xyu = x + u + v for x. The fact that there are letters (rather than numbers) multiplied by x is essentially irrelevant.
Here s the equation to be solved: xc + yc + xyu = x + u + v Move terms with x to the left side, and terms without x to the right side: xc + xyu x = u + v yc Factor out x from terms on the left: x(c + yu ) = u + v yc Divide both sides by c + yu to find the solution : x = u + v yc c + yu More examples of this type: SRW Examples 1.5.1, 1.5., and 1.5.3 on page 46. Things get harder when the equation is not linear. Procedure To solve a nonlinear equation in x (degree or more) Rewrite the equation in the form p(x) = 0 Factor p(x) completely. Set each factor to zero and solve for x.
Example 4a: Solve x 3 = x as follows: Rewrite as x 3 x = 0. Factor: x(x 1) = 0, and so x(x 1)(x + 1) = 0. Set each factor to 0 and solve for x. x = 0 has solution x = 0; x 1 = 0 has solution x = 1; and x + 1 = 0 has solution x = 1. Answer: the solutions of x 3 = x are 0, 1, and 1. Warning: never solve an equation by guessing. If you say that the solution of x = x is x = 1, you receive zero part credit. Example 4b: Solve x 4 = 4x 3 3x as follows: Rewrite as x 4 4x 3 + 3x = 0. Factor: x (x 4x + 3) = x (x 3)(x 1). Set each factor to 0 and solve for x. x = 0 has solution x = 0; x 3 = 0 has solution x = 3; and x 1 = 0 has solution x = 1. Answer: the solutions of x 4 = 4x 3 3x are 0, 1, and 3.
Quadratic equations Definition A quadratic equation is a degree polynomial equation. We will always assume that the equation is written as ax + bx + c = 0. with a 0. The easiest quadratic equation to solve is x = K. This is the special case of ax + bx + c = 0 with a = 1, b = 0, c = K. If K > 0, the equation x = K has two solutions: x = K and x = K. If K = 0, the equation x = K has one solution: x = 0. If K < 0, the equation x = K has no (real) solution. You get zero part credit if you say the solution of x = 4 is x =. The above statements are true if any expression is substituted for x. To solve (x + 4) = 3, set x + 4 = 3 and x + 4 = 3, to obtain x = 4 + 3 andx = 4 3. We usually write this more briefly as x = 4 ± 3. In general, however, b will not be 0 in the quadratic equation ax + bx + c = 0. In this section we use two methods to solve this equation:
Method 1: Factor ax + bx + c and set each factor to 0. Method : Use the quadratic formula. Method 1 is the method described above for solving polynomial equations. In some cases, the factoring is difficult, and it is easier to use Method. Example 5: Solve x(x + 1) = 1 Method 1 Solution: Rewrite as x + x = 1, then as x + x 1 = 0 Factor x + x 1 = as (x + r)(x + s) by solving r + s = 1, rs = 1 to get: x + x 1 = (x + 4)(x 3) = 0 Set each factor to 0 and solve forx: x + 4 = 0 has solution x = 4. x 3 = 0 has solution x = 3. Answer: the solutions of x(x + 1) = 1 are x = 4 and x = 3. In this problem x + x 1 was easy to factor. However, if the problem changed slightly, say to x + x 10 = 0, then using the factoring method would be a waste of time, since the quadratic does not factor. We will discuss below a method for deciding in advance whether a quadratic polynomial factors.
In contrast, the quadratic formula is a sure thing. Procedure Quadratic Formula : The solutions of the quadratic equation ax + bx + c = 0 are x = b ± b 4ac. The discriminant in this formula is D = b 4ac. a If D > 0, there are two distinct real number solutions; If D = 0, there is one real solution; If D < 0, there are two distinct complex number solutions. Example 6: Use the quadratic formula to solve x + x 1 = 0. Solution: Here a = 1, b = 1, c = 1, and so x = b± b 4ac a = 1± 1 4(1)( 1) (1) = 1± 49 = 1±7 Then the solutions are x = 1 7 = 8 = 4 and x = 1+7 = 6 = 3. In this problem, factoring was easier than the quadratic formula. Sometimes the reverse is true. Exercise: Solve 6x + 13x + 6 = 0, first by trial and error factoring, and then by using the quadratic formula.
While the quadratic formula is a sure thing, there are situations where you need to know in advance whether a quadratic polynomial factors. For example, suppose you are asked to simplify (reduce to lowest terms) the fraction 6x + 13x + 6. You need to 3x + decide if the numerator factors. There is a straightforward method of deciding in advance whether it is possible to factor a quadratic polynomial. The discriminant D = b 4ac tells you the whole story, as follows. Quadratic Polynomial Factoring Criterion: The quadratic polynomial ax + bx + c factors if and only if b 4ac is a whole number. Example 7: Show that 6x + 13x + 6 can be factored. Solution: D = b 4ac = 13 4(6)(6) = 169 144 = 5. Since 5 = 5 is a whole number, 6x + 13x + 6 can be factored.
Factoring 6x + 13x + 6 by the usual trial and error method is unpleasant: you need to try a lot of possibilities. Here is a better way. Procedure for factoring a quadratic polynomial. To factor ax + bx + c Calculate D = b 4ac If D is a whole number: Let r 1 and r be the roots of ax + bx + c = 0. Then ax + bx + c = a(x r 1 )(x r ) can be rewritten as a factorization with integer coefficients If D is a not a whole number: there is no factorization with integer coefficients. Example 8: Factor 6x + 13x + 6. In the previous example we saw that b 4ac = 5 = 5 is a whole number. Therefore, the above result guarantees that 6x + 13x + 6 factors with integer coefficients and gives a method of doing the factoring, as follows.
r 1 = 13+ 5 1 = 13+5 1 = 3 ; r = 13 5 1 = 13 5 1 = 3 and so 6x + 13x + 6 = a (x r 1 ) (x r ) = 6 ( x + ( ) 3) x + 3 = 3 ( x + ( 3) x + 3 ) = (3x + )(x + 3) Now read SRW Example 1.5.9, and do the suggested homework exercise. Now we can do the following problem in a systematic way: Example 9: Reduce the fraction 18x + 39x + 18 to lowest terms. 1x + 8 Solution: Carry out the three steps for reducing a fraction to lowest terms: Factor the numerator: 18x + 39x + 18 = 3(6x + 13x + 6) = 3(3x + )(x + 3) Factor the denominator: 1x + 8 = 4(3x + ) Cancel common factors: 18x + 39x + 18 9x + 6 = 3 (3x + )(x + 3) 4 (3x + = ) 3(x + 3) 4
It is important to understand that the coefficients a, b, c in the quadratic equation can be numbers, letters, or any expressions whatsoever. The equation is quadratic in x provided that the only powers of x are x (with non-zero coefficient) and (possibly) x. Example 10: Use the quadratic formula to solve 3x + 4x + stx + q = 0 Solution: Rearrange to get (3 + st)x + 4x + q = 0 This is obtained from the quadratic equation ax + bx + c = 0 by substituting 3 + st,for a, 4 for b, and q for c. Answer: The solutions are x = b ± b 4ac a = 4 ± 4 4(3 + st)(q) (3 + st) So far, we have discussed how to solve polynomial equations, which do not involve fractions. In equations that do involve fractions, we simply rewrite them as polynomial equations. Definition A rational equation is one in which both sides involve polynomials or quotients of polynomials. Read Example SRW 1.5.10
Rational equations Procedure To solve a rational equation Multiply both sides by the LCD of all denominators that appear. Solve the resulting polynomial equation. Reject any solutions that are roots of any denominator of the original equation. 4 Example 11: Solve the rational equation x = + 4 3 x Solution: here is the original problem 4 x = + 4 3 x Denominators ae prime polynomials LCD = x(3 x) Multiply both sides by LCD x(3 x) ( ) ( ) 4 x = x(3 x) + 4 3 x Distribute the LCD and cancel (3 x)4 = x(3 x)() + 4x Remove parentheses 1 4x = (3x x ) + 4x to get 1 4x = 6x x + 4x a quadratic equation x 14x + 1 = 0
Divide by x 7x + 6 = 0 Factor (x 6)(x 1) = 0 Set each factor to zero x = 6 and x = 1 are possible solutions. Check the answer x = 1 by substituting 1 for x in the original equation 4 Original equation x = + 4 3 x 4 Substitute 1 for x 1 =? + 4 3 1 4 =? + 4 4 =? + Yes! Check the answer x = 6 by substituting 1 for x in the original equation 4 Original equation x = + 4 3 x 4 Substitute 6 for x 6 =? + 4 3 6 Rewrite 3 =? + 4 3 6 3 =? 6 3 4 3 Yes! Answer: The equation 4 x = + 4 3 x has solutions x = 1 and x = 6. Alse see Example SRW 1.5.10 and do the suggested exercise.
Equations with radicals The last category of equations we will discuss involves radical signs, typically square roots. To solve these, we get rid of the square roots and then solve the remaining polynomial or rational equation: To solve an equation involving one or more radicals: Rearrange the equation so that a radical is alone on the left side. Square both sides of the equation. Solve the resulting polynomial equation. Check your solution(s) in the original equation. Remember that any solution involving the square root of a negative number must be rejected. Example 1: Solve x + 1 + 1 = x Solution: Original equation x + 1 + 1 = x Isolate radical on left x + 1 = x 1
Square both sides ( ) x + 1 = (x 1) x + 1 = x x + 1 Rewrite equation in x = x x standard form and factor 0 = x 4x = x(x 4) to get possible solutions x = 0 and x = 4. Check the answer x = 0 by substituting 0 for x in the original equation Original equation x + 1 + 1 = x Substitute 0 for x 0 + 1 + 1 =? 0 and check 1 + 1 =? 0. No! Reject x = 0 Check the answer x = 4 by substituting 4 for x in the original equation Original equation x + 1 + 1 = x Substitute 0 for x (4) + 1 + 1 =? 4 9 + 1 =? 4 and check 3 + 1 =? 4. Yes! Accept the answer x = 4. Answer: The equation x + 1 + 1 = x has one solution x = 4 Read Example SRW 1.5.11 and do the suggested exercise.
Substituting a variable for a power of the unknown Example 13: Solve x 4 8x + 8 = 0 Solution: Notice that the only powers of x are x and x 4. This suggests rewriting x 4 8x + 8 = 0 as (x ) 8x + 8 = 0. Substitute W for x to get W 8W + 8 = 0. Now apply the quadratic formula: W = 8± ( 8) 4 1 8 = 8± 3 = 8±4 = 4 ± Thus there are two solutions for W, namely W 1 = 4 + and W = 4. Since these are both positive and x = W, the original equation has four solutions: x = ± W 1 = ± 4 + and x = ± W = ± 4 Example 14: Solve x 1/3 + x 1/6 = 0. Solution: Rewrite as (x 1/6 ) + x 1/6 = 0. Setting W = x 1/6 yields W + W ) = (W + )(W 1) = 0 and so W = or W = 1. Since x = W 6, we get x = ( ) 6 = 64 or x = 1 6 = 1. Check your answer by substituting in the original equation. x = 1 is fine, but x = 64 yields the false statement 4 + = 0. Solution: x = 1.
Exercises for Chapter 1.5: Equations Click on Wolfram Calculator to find an answer checker. Click on Wolfram Algebra Examples to see how to check various types of algebra problems. 1. Do the WebAssign exercises.. Solve each of the following equations: a) x = x b) x(x + 1) = 0 c) x 3 = 16x d) x 4 = 16x e) x = 0 f) x(x + 1)(x + ) = (x + 1)(x + 5x + 6) g) 1x + 30x = 5x 1 3. Solve each of the following equations: a) x + 4 = 0 b) x 3 + x x = 0 c)x 4 4 = 0 4. Solve each of the following equations: a) x 4/3 5x /3 4 + 6 = 0 b) x + 1 x = 108 c) 3x + 7 = x 7 x d) 10 x 1 x 3 + 4 = 0 e) x x+7 x+1 x+3 = 1 f) 5 x + 1 = x
Quiz Review Example 1: Find the degree of each equation: 1 + x + x + x 3 = x + 7 x + 3x = x 5 Example :: Solve 3x + 5 = 8x. Example 3:: Solve xc + yc + xyu = x + u + v for x. Example 4: Solve each of the following x 3 = x x 4 = 4x 3 3x Example 5: Solve x(x + 1) = 1 Example 7: How do you know that 6x + 13x + 6 can be factored? Example 8: Factor 6x + 13x + 6. Example 9: Reduce the fraction 18x + 39x + 18 to lowest terms. 1x + 8 Example 10: Use the quadratic formula to solve 3x + 4x + stx + q = 0. 4 Example 11: Solve the rational equation x = + 4 3 x. Example 1: Solve x + 1 + 1 = x. Example 13: Solve x 4 8x + 8 = 0. Example 14: Solve x 1/3 + x 1/6 = 0.