3.2 Solving Quadratic Equations

Transcription

1 Section 3. Solving Quadratic Equations 1 3. Solving Quadratic Equations Let s begin with a review of an important algebraic shortcut called squaring a binomial. Squaring a Binomial A binomial is an algebraic sum or difference of two monomials (or terms), such as x + y or 3ab c 3. To square a binomial, start with an arbitrary binomial, such as a + b, then multiply it by itself to produce its square (a + b)(a + b), or, more compactly, (a + b). Algorithm for Squaring a Binomial. To square the binomial a + b, proceed as follows: 1. Square the first term to get a.. Multiply the first and second terms together, and then multiply the result by two to get ab. 3. Square the second term to get b. Thus, to expand (x + 4), we should proceed as follows. 1. Square the first term to get x. Multiply the first and second terms together and then multiply by two to get 8x. 3. Square the second term to get 16. Proceeding in this manner allows us to perform the expansion mentally and simply write down the solution. (x + 4) = x + (x)(4) + 4 = x + 8x + 16 Here are a few more examples. In each, we ve written an extra step to help clarify the procedure. In practice, you should simply write down the solution without any intermediate steps. (x + 3) = x + (x)(3) + 3 = x + 6x + 9 (x 5) = x + (x)( 5) + ( 5) = x 10x + 5 ( x 1 ) = x + (x) ( 1 ) ( + 1 ) = x x Perfect Square Trinomials Once you ve mastered squaring a binomial, as in (a + b) = a + ab + b, (3.1) it s a simple matter to identify and factor trinomials (three terms) having the form a + ab + b. You simply undo the multiplication. Whenever you spot a trinomial

2 Chapter 3 whose first and third terms are perfect squares, you should suspect that it factors as follows. a + ab + b = (a + b) (3.) A trinomial that factors according to this rule or pattern is called a perfect square trinomial. For example, the first and last terms of the following trinomial are perfect squares. x + 16x + 64 The square roots of the first and last terms are x and 8, respectively. Hence, it makes sense to try the following. x + 16x + 64 = (x + 8) It is important that you check your result using multiplication. So, following the three-step algorithm for squaring a binomial: 1. Square x to get x.. Multiply x and 8 to get 8x, then multiply this result by to get 16x. 3. Square 8 to get 64. Hence, x + 16x + 64 is a perfect square trinomial and factors as (x + 8). As another example, consider x 10x + 5. The square roots of the first and last terms are x and 5, respectively. Hence, it makes sense to try x 10x + 5 = (x 5). Again, you should check this result. Note especially that twice the product of x and 5 equals the middle term on the left, namely, 10x. Zeros of the Quadratic We ve reviewed how vertex form and the axis of symmetry can help to draw an accurate graph of the quadratic function defined by the equation f(x) = ax + bx + c. When drawing the graph of the parabola it is also helpful to know where the graph of the parabola crosses the x-axis. We ll review the techniques that will enable you to factor the quadratic expression ax + bx + c which will assist in solving ax + bx + c = 0. Factoring ax + bx + c when a = 1 Let s look at an example. Example 1 Factor the trinomial x 5x 84. List all the integer pairs whose product is 84.

3 Section 3. Solving Quadratic Equations 3 1, 84 1, 84, 4, 4 3, 8 3, 8 4, 1 4, 1 6, 14 6, 14 7, 1 7, 1 We ve framed the pair whose sum equals the coefficient of x, namely 5. Use this pair to factor the trinomial. x 5x 84 = (x + 3)(x 8) Check. (x + 3)(x 8) = x(x 8) + 3(x 8) = x 8x + 3x 84 = x 5x 84 Now, let s review how to proceed when the leading coefficient is not 1. Factoring ax + bx + c when a 1 When a 1, we use a technique called the ac-test to factor the trinomial ax + bx + c. The process is best explained with an example. Example Factor x + 13x 4. Note that the leading coefficient does not equal 1. Indeed, the coefficient of x in this example is a. Therefore, the technique of the previous examples will not work. Thus, we turn to a similar technique called the ac-test. First, compare x + 13x 4 and ax + bx + c and note that a =, b = 13, and c = 4. Compute the product of a and c. This is how the technique earns its name ac-test. ac = ()( 4) = 48 List all integer pairs whose product is ac = 48.

4 4 Chapter 3 1, 48 1, 48, 4, 4 3, 16 3, 16 4, 1 4, 1 6, 8 6, 8 We ve framed the pair whose sum is b = 13. The next step is to rewrite the trinomial x + 13x 4, splitting the middle term into a sum, using our framed integer pair. x + 13x 4 = x 3x + 16x 4 We factor an x out of the first two terms, then an 8 out of the last two terms. This process is called factoring by grouping. x 3x + 16x 4 = x(x 3) + 8(x 3) We now factor out a common factor of x 3. x(x 3) + 8(x 3) = (x + 8)(x 3) It s helpful to see the complete process as a coherent unit. x + 13x 4 = x 3x + 16x 4 = x(x 3) + 8(x 3) = (x + 8)(x 3) Check. Again, it is important to check the answer by multiplication. (x + 8)(x 3) = x(x 3) + 8(x 3) = x 3x + 16x 4 = x + 13x 4 Because this is the original trinomial, our solution checks. Intercepts Recall the process for finding the x-intercepts. Finding x-intercepts. To find the x-intercepts of the graph of any function, set y = 0 and solve for x. Alternatively, if function notation is used, set f(x) = 0 and solve for x. Let s look at an example. Example 3 Find the x-intercepts of the graph of the quadratic function defined by y = x +x 48.

5 Section 3. Solving Quadratic Equations 5 To find the x-intercepts, first set y = 0. 0 = x + x 48 Next, factor the trinomial on the right. Note that the coefficient of x is 1. We need only think of two integers whose product equals the constant term 48 and whose sum equals the coefficient of x, namely. The numbers 8 and 6 come to mind, so the trinomial factors as follows (readers should check this result). 0 = (x + 8)(x 6) To complete the solution, we need to use an important property of the real numbers called the zero product property. Zero Product Property. If a and b are any real numbers such that then either a = 0 or b = 0. ab = 0, In our case, we have 0 = (x + 8)(x 6). Therefore, it must be the case that either x + 8 = 0 or x 6 = 0. These equations can be solved independently to produce x = 8 or x = 6. Thus, the x-intercepts of the graph of y = x +x 48 are located at ( 8, 0) and (6, 0). Let s look at another example. Example 4 Find the x-intercepts of the graph of the quadratic function f(x) = x 7x 15. To find the x-intercepts of the graph of the quadratic function f, we begin by setting f(x) = 0. Of course, f(x) = x 7x 15, so we can substitute to obtain x 7x 15 = 0. We will now use the ac-test to factor the trinomial on the left. Note that ac = ()( 15) = 30. List the integer pairs whose products equal 30.

6 6 Chapter 3 1, 30 1, 0, 15, 15 3, 10 3, 10 5, 6 5, 6 Note that the framed pair sum to the coefficient of x in x 7x 15. Use the framed pair to express the middle term as a sum, then factor by grouping. x 7x 15 = 0 x + 3x 10x 15 = 0 x(x + 3) 5(x + 3) = 0 (x 5)(x + 3) = 0 Now we can use the zero product property. Either x 5 = 0 or x + 3 = 0. Each of these can be solved independently to obtain x = 5 or x = 3/. Thus, the x-intercepts of the graph of the quadratic function f(x) = x 7x 15 are located at ( 3/, 0) and (5, 0). Zeros of a Function Recall the difference between the zeros of a function and the x intercepts. Definition 1 Zeros of a Function. function f. The solutions of f(x) = 0 are called the zeros of the Thus, in the last example, both 3/ and 5 are zeros of the quadratic function f(x) = x 7x 15. Note the intimate relationship between the zeros of the quadratic function and the x-intercepts of the graph. Note that 3/ is a zero and ( 3/, 0) is an x-intercept. Similarly, 5 is a zero and (5, 0) is an x-intercept. The Difference of Two Squares A special multiplication pattern that appears frequently is called the difference of two squares. Use the distributive property to expand (a + b)(a b). (a + b)(a b) = a(a b) + b(a b) = a ab + ba b

7 Section 3. Solving Quadratic Equations 7 Since ab = ba, we have the following result. Property 1 The Difference of Two Squares Pattern: (a + b)(a b) = a b Recall that this pattern is used in factoring polynomials of this form. Let s look at an example. 4x 9 = (x + 3)(x 3). We start by taking the square root of the two squares. Thus, the square root of 4x is x and the square root of 9 is 3. We then form two binomials with the results x and 3 as matching first and second terms, separating one pair with a plus sign, the other pair with a minus sign. In similar fashion, Quadratic Formula 9x 49 = (3x + 7)(3x 7). Many times we have a quadratic equation that we would like to solve and it doesn t factor easily. Recall the following formula for solving any quadratic equation. The Quadratic Formula. The solutions to the quadratic equation ax + bx + c = 0 a > 0 (3.3) are given by the quadratic formula x = b ± b 4ac. (3.4) a Let s look at some examples. Example 5 Use the quadratic formula to solve the equation x = 7 6x. The first step is to place the equation in the form ax + bx + c = 0 by moving every term to one side of the equationarranging the terms in descending powers of x. x + 6x 7 = 0

8 8 Chapter 3 Next, compare x + 6x 7 = 0 with the general form of the quadratic equation ax + bx + c = 0 and note that a = 1, b = 6, and c = 7. Copy down the quadratic formula. x = b ± b 4ac a Substitute a = 1, b = 6, and c = 7 and simplify. x = (6) ± (6) 4(1)( 7) (1) x = 6 ± x = 6 ± 144 In this case, 144 = 1, so we can continue to simplify. x = 6 ± 1 It s important to note that there are two real answers, namely Simplifying, x = 6 1 or x = x = 9 or x = Let s look at another example. Example 6 Given the quadratic function f(x) = x x, find all real solutions of f(x) =. Because f(x) = x x, the equation f(x) = becomes x x =. Set one side of the equation equal to zero by subtracting from both sides of the equation. x x = 0 Compare x x = 0 with the general quadratic equation ax + bx + c = 0 and note that a = 1, b = and c =. Write down the quadratic formula. x = b ± b 4ac a Next, substitute a = 1, b =, and c =. Note the careful use of parentheses.

9 Section 3. Solving Quadratic Equations 9 Simplify. x = ( ) ± ( ) 4(1)( ) (1) x = ± x = ± 1 In this case, 1 is not a perfect square, so we ve simplified as much as is possible at this time. The Discriminant Consider again the quadratic equation ax + bx + c = 0 and the solutions (zeros) provided by the quadratic formula x = b ± b 4ac. a The expression under the radical, b 4ac, is called the discriminant, which we denote by the letter D. That is, the formula for the discriminant is given by D = b 4ac. The discriminant is used to determine the nature and number of solutions to the quadratic equation ax +bx+c = 0. This is done without actually calculating the solutions. Let s look at three key examples. Example 7 Consider the quadratic equation x 4x 4 = 0. Calculate the discriminant and use it to determine the nature and number of the solutions. The discriminant is given by the calculation D = b 4ac = ( 4) 4(1)( 4) = 3. Note that the discriminant D is positive; i.e., D > 0. If we graph f(x) = x 4x 4 we see that the function has two x-intercepts (found using the quadratic formula). Thus, if the discriminant is positive, the parabola will have two real x-intercepts.

10 10 Chapter 3 y ( 4 3 ) ( ), , 0 x (, 8) Figure 1. The graph of f(x) = x 4x 4. Next, let s look at an example where the discriminant equals zero. Example 8 Consider the quadratic equation x 4x + 4 = 0. Calculate the discriminant and use it to determine the nature and number of the solutions. The discriminant is given by the calculation D = b 4ac = ( 4) 4(1)(4) = 0. Note that the discriminant equals zero. If we write f(x) = x 4x + 4 in vertex form we can determine the vertex by inspection. f(x) = (x ). (3.5) This is a parabola that opens upward and is shifted units to the right. Note that there is no vertical shift, so the vertex of the parabola will rest on the x-axis, as shown in Figure. The key thing to note here is that the discriminant D = 0 and the parabola has only one x-intercept. That is, the equation x 4x + 4 = 0 has a single real solution. Next, let s look what happens when the discriminant is negative. Example 9 Consider the quadratic equation x 4x + 8 = 0.

11 Section 3. Solving Quadratic Equations 11 y 10 (, 0) x 10 x = Figure. The graph of f(x) = (x ). Calculate the discriminant and use it to determine the nature and number of the solutions. The discriminant is given by the calculation D = b 4ac = ( 4) 4(1)(8) = 16. Note that the discriminant is negative. If we put our function in vertex form and graph it we see that it does not cross the x axis. y 10 (, 4) x 10 x = Figure 3. The graph of f(x) = (x ) + 4. The key point in this example is the fact that the discriminant is negative and there are no real solutions of the quadratic equation (equivalently, there are no x-intercepts). If we try to find the solutions of x 4x + 8 = 0 we ll see that there are no real solutions. Let s summarize what we have reviewed.

12 1 Chapter 3 Summary 1 Consider the quadratic function f(x) = ax + bx + c. The graph of this function is a parabola. Three possibilities exist depending upon the value of the discriminant D = b 4ac. 1. If D > 0, the parabola has two x-intercepts and therefore two zeros.. If D = 0, the parabola has exactly one x-intercept and therefore one zero. 3. If D < 0, the parabola has no x-interceptsand therefore no zeros. Notice in our previous examples that we could have determined the number of and type of solutions using the discriminant.