Martin-Gay, Developmental Mathematics 1 Solving Quadratic Equation by Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then Example ♦ Solve x2 = 49 x 49 7 ♦ Solve 2x2 = 4 x2 = 2 x 2 ♦ Solve (y – 3)2 = 4 y 3 4 2 y=32 y = 1 or 5 a b ♦ Solve x2 + 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number. Example Solve (x + 2)2 = 25 x 2 25 5 x = 2 ± 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7 Solve (3x – 17)2 = 28 3x – 17 = 28 2 7 3x 17 2 7 17 2 7 x 3 Solving quadratic Equation by Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples). Example What constant term should be added to the following expressions to create a perfect square trinomial? x2 – 10x add 52 x2 – 7x = 25 x2 + 16x add 82 = 64 2 7 49 add 4 2 Example We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Solving a Quadratic Equation by Completing a Square 1) If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. 2) Isolate all variable terms on one side of the equation. 3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation). 4) Factor the resulting trinomial. 5) Use the square root property. Example Solve by completing the square. y2 + 6y + 9 = 8 + 9 (y + 3)2 = 1 y2 + 6y = 8 y + 3 = ± 1= ± 1 y = 3 ± 1 y = 4 or 2 Solve by completing the square. y2 + y = 7 y2 + y + ¼ = 7 + ¼ 29 (y + ½)2 = 4 1 29 29 y 2 4 2 1 29 1 29 y 2 2 2 y2 + y – 7 = 0 The Quadratic Formula ♦ Another technique for solving quadratic equations is to use the quadratic formula. ♦ The formula is derived from completing the square of a general quadratic equation. ♦ A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions. Example b b 2 4ac x 2a ♦ Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1 9 (9) 2 4(11)(1) 9 81 44 9 125 9 5 5 n 22 22 22 2(11) The Discriminant ♦The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. ♦The discriminant will take on a value that is positive, 0, or negative. ♦The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively. Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions. Steps in Solving Quadratic Equations 1. 2. 3. 4. If the equation is in the form (ax + b)2 = c, use the square root property to solve. If not solved in step 1, write the equation in standard form. Try to solve by factoring. If you haven’t solved it yet, use the quadratic formula. Example ♦ Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = -3, c = 1 ♦ Solve the following 5 2 1 m m 0 8 2 5m2 8m 4 0 3 (3) 2 4(1)(1) x 2(1) (5m 2)(m 2) 0 3 9 4 3 5 2 2 5m 2 0 or m 2 0 2 m or m 2 5 Graphs of Quadratic Equations ♦The graph of a quadratic equation is a parabola. ♦The highest point or lowest point on the parabola is the vertex. Example Graph x y= 2x2 y – 4. (–2, 4) y 2 4 1 –2 0 –4 –1 –2 –2 4 (2, 4) x (–1, – 2) (1, –2) (0, –4) Intercepts of the Parabola Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y. Characteristics of the Parabola ♦ If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. 2) the x-coordinate of the vertex is b . 2a To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y. Example Graph Since a = –2 and b = 4, the graph opens down and the xcoordinate of the vertex is 4 1 2(2) x y 3 –1 2 5 1 7 0 5 –1 –1 y = –2x2 + 4x + 5. y (0, 5) (–1, –1) (1, 7) (2, 5) (3, –1) x Domain and Range Recall that a set of ordered pairs is also called a relation. The domain is the set of x-coordinates of the ordered pairs. The range is the set of y-coordinates of the ordered pairs. Example Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)} ♦ Domain is the set of all x-values, {4, –4, 2, 10} ♦ Range is the set of all y-values, {9, 3, –5} Example Find the domain and range of the function graphed to the right. Use interval notation. y Domain x Domain is [–3, 4] Range is [–4, 2] Range Find the domain and range of the function graphed to the right. Use interval notation. y Range x Domain is (– , ) Range is [– 2, ) Domain Graphing Piecewise-Defined Functions Example 3x 2 if x 0 Graph f (x) . x 3 if x 0 Graph each “piece” separately. Values 0. x f (x) = 3x – 1 x f (x) = x + 3 0 – 1(closed circle) 1 4 2 5 3 6 –1 – 4 –2 – 7 Values > 0. Continued. Example continued y x f (x) = 3x – 1 0 – 1(closed circle) –1 – 4 –2 – 7 (3, 6) Open circle (0, 3) (0, –1) x f (x) = x + 3 1 4 2 5 3 6 (–1, 4) (–2, 7) x The End Martin-Gay, Developmental Mathematics 18

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