Solving Quadratic & Higher Degree Equations

Chapter 7 Solving Quadratic & Higher Degree Equations Sec 1. Zero Product Property Back in the third grade students were taught when they multiplied a number by zero, the product would be zero. In algebra, that s an extremely important property, so important it has a name, the Zero Product Property. While teachers may not have used that name, the point they tried to get across was that the product of any number and zero is zero. 5 x 0 = 0, 123 x 0 = 0, and 0 x 4.32 = 0. It s a pretty straight forward concept. Extending the thinking, we found that in order for the product of 4 and some number n to be zero, n would have to be zero. 4n = 0 n = 0 If I multiplied two numbers, x and y, and found their product to be zero, then one of those two numbers would have to be zero. Mathematically, we d write xy = 0 then x = 0 or y = 0. One of the two numbers would be equal to zero. But, there is another possibility, both x and y could be equal to zero. So a list of all possible answers would include, x = 0, y = 0, or x = 0 and y = 0. An application of the Zero Product Property will allow us to solve higher degree equations by changing them into multiplication problems whose product is zero. For example, if I were asked to solve, x 2 x = 12, I might have a little trouble. If I changed the problem into a multiplication problem (factoring), I would have x(x 1) = 12 I have two numbers multiplied together that equal 12. Through a trial and error process, I might find values of x that would satisfy the original equation. But, if I used the Zero Product Property, I would be able to solve the problem using the logic we used in third grade the product is zero if I multiplied by zero. Rewriting x 2 x = 12 to x 2 x 12 = 0, then factoring, I would have the product of two numbers equaling zero rather than 12. By using the third grade logic, I know that one of the two numbers or maybe both of them are zero since the product is zero. In other words, we would have x 2 x 12 = 0 equivalent to (x 4)(x + 3) = 0. Copyright 2003 Hanlonmath 1

I have a product of two numbers equaling zero, that means one of them has to be zero. When is x 4 = 0? When is the other number x + 3 = 0? When you answer those questions, you have found the values of x that make the open sentence true. Simply put, you solved the equation. The answer, the solution, the zeros are x = 4 or x = -3. The reason we learned how to factor is so we could solve Quadratic and higher degree equations. Those skills are important. If you are most comfortable factoring, then do more problems. You not only need that skill for solving the following quadratic equations, you ll also need to factor for adding and subtracting rational expressions and reducing. A quadratic equation is an equation of degree 2, that is, the exponent on the variable is 2. EXAMPLE: 3x² + 5x 4 = 0 There are a number of different methods that can be used for solving quadratic equations, we ll look at two of these methods. We ll solve them by FACTORING and the QUADRATIC FORMULA. Let s make sure we know what factoring is; FACTORING is the process of changing an expression that is essentially a sum into an expression that is essentially a product. Depending upon what the expression looks like, we may factor using the DISTRIBUTIVE PROPERTY, DIFFERENCE OF 2 SQUARES, LINEAR COMBINATION, TRIAL AND ERROR, GROUPING AND THE SUM OR DIFFERENCE OF TWO CUBES. It is very, very important that you can recognize these patterns for factoring. EXAMPLE: Factor: x² + 7x + 12 = 0 We want to change this addition problem into a multiplication problem by factoring (x + 4) (x + 3) = x² + 7x +12 Now, to solve QUADRATIC EQUATIONS by FACTORING we ll use this strategy. 1. Put everything on one side of the equal sign, zero on the other side. 2. Factor the expression completely. 3. Set each factor equal to zero. 4. Solve the resulting equations. EXAMPLE: Solve x² 35 = 2x 1. Put everything on one side, zero on the other side. x² 2x 35 = 0 Copyright 2003 Hanlonmath 2

2. Factoring (x + 5) (x 7) = 0 3. Set each factor equal to zero x + 5 = 0 x 7 = 0 4. Solve x = -5, x = 7 Let s look at the reason why this method works. By putting everything on one side and factoring, we are looking for two numbers when multiplied together equal zero. If you think for a minute, you ll remember the only time you can get an answer of zero in a multiplication problem is when one or both of the original numbers is zero. That is such an important concept in math that we give it a name; the Zero Product Property. It states if ab = 0, then a = 0 or b = 0 or both a and b are zero. EXAMPLE: 8y² + 2y = 3 1. Put everything on one side 8y² + 2y 3 = 0 2. Factor (4y + 3) (2y 1) = 0 3. Use the Zero Product Property 4y + 3 = 0 or 2y 1 = 0 4. Solve: 4y = -3 2y = 1 y = -¾ or y = ½ Knowing what factoring is and being comfortable with the different patterns involved in factoring will either make or break you with quadratic equations. Practice with factoring is essential. If we had a cubic equation to solve, factoring is an appropriate method to use. EXAMPLE: x³ + 12x = 7x² Putting everything on the side we have: x³ - 7x² + 12x = 0 Factoring x(x² - 7x + 12) = 0 x(x 3) (x 4) = 0 Now I have three numbers multiplied together that equals zero. That means at least one of those numbers; x, (x 3) or (x 4) must be zero. That s an extension of the Zero Product Property. Setting each factor equal to zero we have: x = 0 x 3 = 0 x 4 = 0 The solutions are x = 0 x = 3 or x = 4 Copyright 2003 Hanlonmath 3

Piece of cake, don t you think? Plug them into the original equation and you see they work. Solve by factoring. x² + 7x + 12 = 0 x² + 9x + 20 = 0 x² + 6x + 5 = 0 x² 5x + 6 = 0 x² x 20 = 0 x² 11x + 30 = 0 x² + 3x 40 = 0 x² 13x + 36 = 0 x² + 15x + 50 = 0 3x² 21x = 0 x² 6x = 0 6x² + 10x 4 = 0 2x² + 5x + 2 = 0 x² + 5x = 4 x² + 9x = 18 x² + 2x = 3 5x² = 40x x² + 12x = 27 Sec. 2 Simplifying Radicals From grade school, you can probably remember how to take square roots of numbers like 25, 64, and the 100. The number on the inside of the radical sign is called the radicand. Finding those answers are easy when the radicand is a perfect square. But what happens if the radicand is not a perfect square? Knowing you were thinking just that very thought, you must be very pleasantly surprised to know I am going to tell you how to simplify those expressions. Let me just tell you how, then I ll use an example to make it more clear. To simplify a square root (second root), you rewrite the radicand as a product of a perfect square and some other number, then you take the square root of the number you know. The number you don t know the square root of stays inside the radical. EX. 75 We don t know the 75 so we ll rewrite the radicand as a product of a perfect square and some other number. Since 25 is a perfect square that is a factor of 75, we ll rewrite. 75 = 25x 3 = 25 x 3 Copyright 2003 Hanlonmath 4

= 5 3 That s pretty straight forward. If you are not familiar with perfect squares, you should write them down. By multiplying the counting numbers by themselves. Examples; 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,. EX 72 72 = 36x 2 = 36 x 2 = 6 2 Now, what would have happened if you rewrote 72 as 9 x 8? Let s see. 72 = 9x 8 = 9 x 8 = 3 8 Oh, oh, that s not the same answer we got before. How can that be? Well, one reason is we re not finished, we can simplify 8. That s the same answer we got before. 3 8 = 3 4x 2 = 3 4 x 2 = 3 x 2 x 2 = 6 2 Whew! You would not have able to go to bed tonight thinking about that. When simplifying radicals, it would be a good idea to become comfortable with perfect squares. And just as importantly, when you rewrite the radicand as the product of a perfect square and some other number, use the LARGEST perfect square you can. Otherwise, you ll end up doing a lot more steps. Copyright 2003 Hanlonmath 5

Remember reducing fractions, you d use the largest factor that divided into the numerator and denominator because that meant you ended up with less work. The same applies here, the larger the perfect square you use, the shorter the problem. Try a couple on your own. 48 175 32 Now, with that subskill out of the way, we are ready to work with the Quadratic Formula. Happy? Sec. 3 Completing the Square Seems like we should finish a picture when we say complete the square. Doesn t It? Remember when we memorized special products back when we were working with polynomials. One of those (a + b)² = a² +2ab + b², a² + 2ab + b² is called a perfect square because it came from squaring (a + b), Using that pattern, let s look at some special products. (x + 3)² = x² + 6x + 9 (x + 5)² = x² + 10x + 25 (x + 10)² = x² + 20x + 100 (x + 4)² = x² + 8x + 16 Now, if we looked at these long enough, we might see some relationships that really don t just jump out at you. But the relationships are important, so I am going to tell you what they are. Copyright 2003 Hanlonmath 6

Looking at the linear term and its relation to the constant, (see how important vocabulary is in math), do you see any way to get from the 6 to the 9 in the first special product from 10 to 25 in the second product? Chances are you don t. (x + 3)² = x² + 6x + 9 If I took half of the 6, and squared it, I d get 9. (x + 5)² = x² + 10x + 25 If I took half the 10, and squared it, I would get 25. Getting excited? Looking at; (x + 10)² = x² + 20x + 100 Taking half of 20 and squaring it gives me 100. Neat, huh. What I see is all those perfect squares have that same property. Now if I ask, Is x² + 6x 11 a perfect square? By taking half of 6 and squaring, you don t get 11. Therefore, it s not a perfect square. Ok, how can we recognize a perfect square? That s important, we can use that information. Let s say I was asked to solve the equation: x² + 6x 11 = 0 I find I can not factor the Trinomial. So It looks like I can t solve it. However, could I transform that equation into a perfect square? Sure I can, otherwise I would not have asked. We know, in order to have a perfect square, the constant term comes from taking half the linear term and squaring it. So, let s push the (-11) out of the way and complete the square. x² + 6x 11 = 0 How do we do that? Put the -11 on the other side of the equation, we have. Copyright 2003 Hanlonmath 7

x² + 6x = 11 Now, let s make a perfect square. Take half of 6 and square it, we get 9. This is important, if we add 9 to the left side, we must add 9 to the right side. x 2 + 6x 11 = 0 x 2 + 6x + _ = 11+ _ x 2 + 6x + 9 = 11+ 9 (x + 3) 2 = 20 x + 3 = ± 20 x = ± 20 3 x = ±2 5 3 We can use the method of completing the square to solve quadratics we can t factor. All we are doing is making an equivalent equation by adding a number to both sides of the equation that will make the polynomial a perfect square. Then we solve the resulting equation. Sec. 4 Quadratic Formula The reason we learned to simplify radicals is because not all polynomials can be factored over the set of Rational Numbers. What that means is that you might be asked to solve a quadratic equation where the polynomial can not be factored. Well, that poses a small problem, but not one we can t handle. If we can t factor it, we can solve the quadratic by Completing the Square we can generalize that method to come up with a formula. So, another method for solving QUADRATIC EQUATIONS is the QUADRATIC FORMULA. That formula is derived from another method of solving quadratic equations, Completing the Square. 2 b ± b 4ac The QUADRATIC FORMULA is x = 2a Rather than write it as one fraction, I am going to write it as two factions Copyright 2003 Hanlonmath 8

2 b b 4ac x = ± 2 a 2a That will eliminate some of the difficulties in reducing algebraic fractions that some of the students experience. You need to memorize that formula, say it once, say it twice, say it 20 times if you have to-but know it! The General Form of a Quadratic Equation looks like this: ax² + bx + c = 0, a is the coefficient of the squared term (quadratic term), b is the coefficient of the x term (linear term) and c is the constant (the number without a variable). Now, all we have to do is substitute the numbers in those positions in the Quadratic Formula. EXAMPLE: 2x² + 3x 5 = 0 In this example a = 2, b = 3, and c = 5 2 b b 4ac Substituting those values in x = ± 2 a 2a we get x = simplifying x = 2 3 4(2)( 5) (3) ± 2(2) 2(2) 3 9 ( 40) ± 4 4 3 49 x = ± 4 4 Therefore one answer is 3 7 x = ± 4 4 3 7 3 7 10 5 + or 1 and the other answer is or =. 4 4 4 4 4 2 To use the QUADRATIC EQUATION, everything must be on one side and zero on the other side, then use those coefficients to plug into the Quadratic Formula. Watch your signs. The Quadratic Formula allows you to solve any Quadratic Equation. The bad news is if you use it to solve an equation, it takes 2 or 3 minutes to do. The good news is to solve a quadratic; it takes 2 to 3 minutes to do. Copyright 2003 Hanlonmath 9

What that means is, if you can see the factors immediately Solve the problem by factoring, its quicker. However, if you don t see the factors and you don t think the polynomial can be factored, use the Quadratic Formula: That way you won t spend forever trying to factor the problem. Try some of these; you should be able to do them both ways: Factoring then using the Zero Product Property and by the Quadratic Formula. 1. x² - 6x 55 = 0 2. x² + 3x = 40 3. 2x² -12 = 5x SOLUTIONS: 1. x = 11 or -5 2. x = -8 or -5 3. x = 3 or 4 2 By use of the quadratic formula solve each of the following equations. Check. Assume R = {all real numbers}. 1. x² 8x + 15 = 0 2. x² + 7x 8 = 0 3. x² + x 42 = 0 4. x² 11x + 30 = 0 5. 2x² x 1 = 0 6. 6x² x 15 = 0 7. 4x² 23x = 6 8. 15x² 16x = 15 9. 8x² 6x = -1 10. 3x² 20x = 7 11. x² 4x + 1 = 0 12. x² + 10x + 21 = 0 13. 4x² 12x + 7 = 0 14. 9x² + 6x 4 = 0 15. x² + 10x + 19 = 0 16. 3x² + 12x + 8 = 0 Sec. 5 Synthetic Substitution Copyright 2003 Hanlonmath 10

The first way you were taught to evaluate a polynomial for a given value of a variable was direct substitution. Simply put, that means you plugged that value into the expression and found the result. EXAMPLE: If P(x) = 3(5) 10 2 2x + 3x 10, find P when x = 5. By substituting, P = 2(5)² + P = 55 The process of direct substitution can become a real pain for higher degree polynomial expressions. There is another method called SYNTHETIC SUBSTITUTION that will make evaluating a polynomial a very simple process. Given some polynomial Q(x) = 3x 3 + 10x² 5x 4 in one variable. You can evaluate Q when x = 2 by plugging in that value as we did before. Q = 3x³ + 10x² - 5x 4 = 3(2)³ + 10(2)² - 5(2) 4 = 3(8) +10(4) 10 4 = 24 + 40 + - 10 4 = 50 So the value of Q is 50 when x is 2. Synthetic substitution is nothing more than a variation the fourth grade division algorithm. You might remember when you divided a polynomial by another like x 2 and there was no remainder, that meant x 2 went into that evenly. If we had graphed it, we would have also noticed that the graph crossed the x-axis at (2,0). If there was a remainder, that remainder would turn out to be the value of the polynomial at that particular value of x an ordered pair. To use SYNTHETIC SUBSTITUTION to evaluate a particular polynomial for a given value of x, we would write the coefficients of Q; Q = 3x³ + 10x² 5x 4 3 10 5 4 Copyright 2003 Hanlonmath 11

Now, we ll leave a space under those coefficients and draw a line. We will also write down the value of the variable to be plugged in. 2 3 10 5 4 Once we do that, we are set up to evaluate Q when x = 2. To accomplish that, we bring down the first number, 3, and multiply by the 2 and place that product (6) under the next number (10). Next, we add those numbers. Now w start again, we multiply that result (16) by the 2, place that under the next coefficient, and add, we get 27. Keep repeating this process. The last value will be the value of Q when x is 2. 2 3 10-5 -4 6 32 54 3 16 27 50 Notice, the last number we got was 50 as we did before when we plugged 2 into Q or found Q(2). EXAMPLE: Evaluate 2x 4 - x³ + 5x + 3 when x = 3 In this example, notice there is no quadratic term, no x². When we write the coefficients, we ll need to write zero for the coefficient of that missing term as a placeholder. In this case, since I want to find the value when x = 3, I bring down the first number 2 and multiply that by 3, I write that result 6 under the next coefficient and add. I now multiply that by the 3, place that result under the next coefficient and add. I continue 3 2-1 0 5 3 6 15 45 150 2 5 15 50 153 Since the last result is 153, the value of that polynomial expression when x = 3 is 153. EXAMPLE Let s evaluate the same expression when x is 3 The process stays exactly the only difference is I write 3 instead of 3. Copyright 2003 Hanlonmath 12

Bring down the first number, multiply by 3, add and continue. 3 2-1 0 5 3-6 21-63 174 2-7 21-58 177 The value of the polynomial expression when x = 3 is 177. Piece of cake. This could be written as the ordered pair, (-3, 177) if we wanted to graph this relation. Let s try another one you say, OK! EXAMPLE: Let y = 2x 4 + x³ - 11x² - 4x 12, find the value of y when x = 2. 2 2 1-11 -4 12 4 10-2 -12 2 5-1 -6 0 The value of y when x = 2 is 0. If we were to write this as an ordered pair, we d have (2, 0). If we were to graph that, we d also notice that (2, 0) is an x-intercept, where a graph crosses the x-axis. In algebra, when we solve quadratic or higher degree equations, we set the equations equal to zero, then find values of the variable that will make the equation true. Notice, in the last example x = 2 made y = 0. Oh wow! What that means is x = 2 is a solution to the equation. 2x 4 + x³ - 11x² - 4x + 12 = 0 In fact, anytime the last number in a synthetic substitution problem is zero, The value we plugged in represents a solution, a zero, or what most of us might call an answer, if we had an equation. Can you feel the excitement running through your veins? Copyright 2003 Hanlonmath 13

If you put this together with the Rational Root Theorem, we now have a very simple method of finding rational roots and if you think about it, another method of factoring. If x = 2 is a zero, a solution, then x 2 must be a factor of the polynomial expression. Now don t you just love how math seems to come together. Use synthetic substitution to find the value of the following polynomials. 1. P(x) = 2x 3 + 3x 2 4x + 5, find P(2) 2. Q(x) = 2x 4 3x + 6, find Q( 1) 3. T(x) = 5x 3 4x 2 + 3x 10, find T(1) 4. H(x) = x 3 x 2 + x 1, find H(5) 5. J(x) = x 4 x 2 + 1 Sec. 6 Rational Root Theorem If P (x) = 0 is a polynomial equation with integral coefficients of degree n in which n a 0 is the coefficients of x, and a n is the constant term, then for any rational root p/q, where p and q are relatively prime integers, p is a factor of a n and q is a factor of a 0 n n 1 x + a1x + + an 1x + a n = 0 a 0 That s math talk. What that means is you have to start with an equation without fractions, and if there are rational answers (roots), the numerator of the answer must be a factor of the constant ( a n ) and the denominator will be a factor of the leading coefficient ( a 0 ). EXAMPLE: Find the rational roots of 2x² + 5x 12 = 0 a 0 = 2 a n = 12 Notice, no fractions, we have integral coefficients. The factors of the leading coefficient 2 are; ± 2, 1 The factors of the constant -12 are; ± 12, 6, 4, 3, 2, 1. Now, we take each of the factors of the constant and put them over each of the factors of the leading coefficient (p/q). All of those fractions will be possible answers (roots) along with their negatives. ± 12 2, 12 1, 6 2, 6 1, 4 2, 4 1, 3 2, 3 1, 2 2, 2 1, 1 2, 1 1. Oh yes you are having fun. Math is your life. Copyright 2003 Hanlonmath 14

The Rational Root Theorem says if there is a rational answer, it must be one of those numbers. Some of those possible answers repeat 6/2 is the same as 3/1 To find which, or if any of those fractions are answer, you have to plug each one into the original equation to see if any of them make the open sentence true. It turns out 3 2 and -4 are solutions. You know that because when you plug either of those numbers into the polynomial, the result is 0. If you used synthetic substitution, the last number would have been 0. If the last number is NOT 0, then the number you are using is not a solution. That s a lot of plugging in. However, if we were able to plug in using SYNTHETIC SUBSTITUTION our work would be much, much easier. One side note, we could have solved this particular equation more efficiently by factoring or by using the Quadratic Formula. The question is, why then am I solving it by the Rational Root Theorem? For practice. Typically you use the Rational Root Theorem when you don t have higher degree equations - not quadratic equations or when you can t factor the polynomial. This is an important theorem because before knowing it, if you could not factor the polynomial, you could not solve the equation. Yes, math is a blast!! EXAMPLE: Find the possible rational solutions of: x³+ 6x² +3x 10 = 0 Can you use the quadratic equation?. Can you factor the polynomials? The answer to both these questions is NO. The only other method we know at this time to solve this equation is by using the Rational Root Theorem. The factors of the leading coefficient are; ± 1 The factors of the constant are; ± 10, 5, 2, 1 Placing the factors of the constant over the factors of the leading coefficient, the possible solutions are: ± 10, 5, 2, 1 Again, to find out if any of these are solutions, I would have to plug them back into the original equation or use synthetic substitution. Copyright 2003 Hanlonmath 15

The Rational Root Theorem does not guarantee that there is a rational solution. So, there are times when none of the possible solutions will work. The equation will have a solution, it just won t be rational. EXAMPLE Determine the possible rational roots of the following equation just by looking. x 4-5x³ + 9x² 7x + 2 = 0 The equation looks longer, but it is not more difficult. The only possible rational solutions are ± 2, 1. Try this one. EXAMPLE Determine the possible rational roots of 2x³ + 3x² 8x + 3 = 0 What are the factors of 2, the leading coefficient? What are the factors of 3, the constant? Therefore, the possible solutions are ± 3 2, 3 1, 1 2, 1 1. Is there a rational solution and if there is, what is it? Rather than plugging in those 8 possible solutions, let s try synthetic substitution. I m going to try whole numbers first only because they are easier to work with than fractions. 1} 2 3 8 3 2 5 3 2 5 3 0 Since the last number is 0, that means that x = 1 is a solution and that x 1 is a factor. As importantly, I now have an equation formed by those coefficients : 2x 2 + 5x 3 = 0 that I can solve using the Rational Root theorem or the Quadratic Formula or factoring. If I solve the reduced equation by factoring, I have 2x 2 + 5x 3 = 0 (2x 1)(x + 3) = 0 So x = ½ or x = 3. The solution set is {1, ½, 3} The Factor and Remainder Theorems are two very closely related theorems; you might want to look them up for the fun of it. Graphically, when you look at Synthetic Substitution, you realize when the last number in Synthetic Substitution is zero, you have a factor, a root a zero. In other words, you know where the graph crosses the x-axis. Copyright 2003 Hanlonmath 16

If the last number is not a zero, then you don t have a root. Using the Remainder Theorem, what you know is the value of the polynomial for that particular number. That can help you locate roots zeros. If you plug in one number and get a positive number at the end, the remainder, then you plug in another number and get a negative number at the end, that would suggest the graph has to cross the x-axis between those two numbers since the you are going from a positive to a negative number. Rather than just checking numbers sequentially, it might be wiser to choose possible rational solutions between those two numbers. That should cut down on your work.` Solve the following equations using the Rational Root Theorem. 1. x 3 + 4x 2 + x 6 = 0 2. x 3 + 2x 2 13x + 10 = 0 3. 2x 3 + 15x 2 + 28x + 15 = 0 4. 3x 3 4x 2 3x 4 = 0 5. x 3 + 7x 2 + 7x 15 = 0 6. 3x 3 x 2 8x _4 = 0 Copyright 2003 Hanlonmath 17