QUADRATIC EQUATIONS The standard form of a quadratic equation A root of a quadratic Solution by factoring Section 2: Completing the square The quadratic formula The discriminant Proof of the quadratic formula Section 3: The graph of  y = A quadratic:  A parabola A QUADRATIC is a polynomial whose highest exponent is 2. ax² + bx + c. The coefficient of x² is called the leading coeffieient. Question 1.  What is the standard form of a quadratic equation? ax² + bx + c = 0. The quadratic is on the left.  0 is on the right. Question 2.  What do we mean by a root of a quadratic? A solution to the quadratic equation. For example, the roots of this quadratic -- x² + 2x − 8 -- are the solutions to x² + 2x − 8 = 0. To find the roots, we can factor that quadratic as (x + 4)(x − 2). Now, if  x = −4, then the first factor will be 0.   While if  x = 2, the second factor will be 0.  But if any factor is 0, then the entire product will be 0.  Therefore, if x = −4 or 2, then x² + 2x − 8 = 0. −4  and  2 are the solutions to the quadratic equation. They are theroots of that quadratic. Conversely, if the roots are a and b say, then the quadratic can be factored as (x − a)(x − b). A root of a quadratic is also called a zero. Because, as we will see, at each root the value of the graph is 0. Question 3.  How many roots has a quadratic? Always two. Because a quadratic (with leading coefficient 1, at least) can always be factored as (x − a)(x − b), and a, bare the two roots. In other words, when the leading coefficient is 1, the root has the opposite sign of the number in the factor.  If (x + q) is a factor, then  x = −q  is a root. −q + q = 0. Problem 1.   If a quadratic can be factored as (x + 3)(x − 1), then what are the two roots? −3 or 1. We say "or," because x can take only one value at a time. Question 4.  What do we mean by a double root? The two roots are equal. The factors will be (x − a)(x − a), so that the two roots are a, a. For example, this quadratic x² − 12x + 36 can be factored as (x − 6)(x − 6). If x = 6, then each factor will be 0, and therefore the quadratic will be 0.  6 is called a double root. When will a quadratic have a double root?  When the quadratic is a perfect square trinomial. Example 1.   Solve for x:   2x² + 9x − 5. Solution.   That quadratic is factored as follows: 2x² + 9x − 5 = (2x − 1)(x + 5). Now, it is easy to see that the second factor will be 0 when x = −5. As for the value of x that will make 2x − 1 = 0,  we must solve that little equation.        We have: 2x = 1 x = 1 2 The solutions are: x = 1 2  or  −5. Problem 2.   How is it possible that the product of two factors ab = 0? Either a = 0 or b = 0. Solution by factoring Problem 3.   Find the roots of each quadratic by factoring.    a)   x² − 3x + 2 b)   x² + 7x + 12 (x − 1)(x − 2) (x + 3)(x + 4) x = 1  or  2. x = −3  or  −4. Again, we use the conjunction "or," because x takes on only one value at a time.    c)   x² + 3x − 10 d)   x² − x − 30 (x + 5)(x − 2) (x + 5)(x − 6) x = −5  or  2. x = −5  or  6.    e)   2x² + 7x + 3 f)   3x² + x − 2 (2x + 1)(x + 3) (3x − 2)(x + 1) x = − 1 2   or  −3. x =  2 3   or  −1.    g)   x² + 12x + 36 h)   x² − 2x + 1 (x + 6)² (x − 1)² x = −6, −6. x = 1, 1. A double root. A double root. Example 2.   c  =  0.   Solve this quadratic equation: ax² + bx  =  0 Solution.   Since there is no constant term -- c  =  0 --  x is a common factor: x(ax + b)   =   0. This implies: x   =   0 or x   =   − b a . Those are the two roots. Problem 4.   Find the roots of each quadratic.    a)   x² − 5x b)   x² + x x(x − 5) x(x + 1) x = 0  or  5. x = 0  or  −1.    c)   3x² + 4x d)   2x² − x x(3x + 4) x(2x − 1) x = 0  or  − 4 3 x = 0  or  ½ Example 3.   b  =  0.  Solve this quadratic equation: ax² − c   =  0. Solution.   In the case where there is no middle term, we can write: ax² = c. This implies: x² = c a x = ,  according to Lesson 26. However, if the form is the difference of two squares -- x² − 16 -- then we can factor it as: (x + 4)(x −4). The roots are ±4. In fact, if the quadratic is x² − c, then we could factor it as: (x + )(x − ), so that the roots are  ±. Problem 5.   Find the roots of each quadratic.    a)   x² − 3 b)   x² − 25 c)   x² − 10 x² = 3 (x + 5)(x − 5) (x + )(x − ) x = ±. x = ±5. x = ±. Example 4.   Solve this quadratic equation: x²  =  x + 20. Solution.   First, rewrite the equation in the standard form, bytransposing all the terms to the left: x² − x − 20  = 0 (x + 4)(x − 5)  = 0 x  = −4  or  5. And so an equation is solved when x is isolated on the left. x = ± is not a solution. Problem 6.   Solve each equation for x.    a)   x²  =  5x − 6 b)   x² + 12  =  8x x² − 5x + 6 = 0 x² − 8x + 12 = 0 (x − 2)(x − 3) = 0 (x − 2)(x − 6) = 0 x = 2  or  3. x = 2  or  6.    c)   3x² + x  = 10 d)   2x²  =  x 3x² + x − 10 = 0 2x² − x = 0 (3x − 5)(x + 2) = 0 x(2x − 1) = 0 x = 5/3  or − 2. x = 0  or  1/2. Example 5.   Solve this equation 3 −  5 2 x − 3x²   =   0 Solution.   We can put this equation in the standard form by changing all the signs on both sides.  0 will not change.  We have the standard form: 3x² +  5 2 x − 3   =   0 Next, we can get rid of the fraction by multiplying both sides by 2.  Again, 0 will not change. 6x² + 5x − 6 = 0 (3x − 2)(2x + 3) = 0. The roots are  2 3  and − 3 2 . Problem 7.   Solve for x.    a)   3 −  11  2 x − 5x²   =  0 b)   4 +  11  3 x − 5x²   =  0 5x² + 11  2 x − 3  =  0 5x² −  11  3 x − 4  =  0 10x² + 11x − 6  =  0 15x² − 11x − 12  =  0 (5x − 2 )(2x + 3)  =  0 (3x − 4)(5x + 3 )  =  0 The roots are  2 5  and − 3 2 . The roots are  4 3  and − 3 5 .    c)    −x² − x + 20  = 0 >  d)     −x² + 3x + 18  = 0 x² + x − 20  = 0 x² − 3x − 18  = 0 (x + 5)(x − 4)  = 0. (x − 6)(x + 3)  = 0. x = −5  or  4. x = 6  or  −3.