Quadratic Equations Solving Quadratic Equations by the Square

Quadratic Equations

Solving Quadratic Equations by the Square Root Property

Square Root Property t. We previously have used factoring to solve quadratic equations. t. This chapter will introduce additional methods for solving quadratic equations. t. Square t. If Root Property b is a real number and a 2 = b, then Martin-Gay, Developmental Mathematics 3

Square Root Property Example t Solve x 2 = 49 Solve 2 x 2 = 4 x 2 = 2 Solve (y – 3)2 = 4 y=3± 2 y = 1 or 5 Martin-Gay, Developmental Mathematics 4

Square Root Property Example t. Solve x 2 +4=0 x 2 = − 4 t. There is no real solution because the square root of − 4 is not a real number. t Martin-Gay, Developmental Mathematics 5

Square Root Property Example t. Solve (x + 2)2 = 25 x = − 2 ± 5 x = − 2 + 5 or x = − 2 – 5 x = 3 or x = − 7 Martin-Gay, Developmental Mathematics 6

Square Root Property Example t. Solve 28 (3 x – 17)2 = 3 x – 17 = Martin-Gay, Developmental Mathematics 7

Solving Quadratic Equations by the Quadratic Formula

The Quadratic Formula t. Another technique for solving quadratic equations is to use the quadratic formula. t. The formula is derived from completing the square of a general quadratic equation. Martin-Gay, Developmental Mathematics 9

The Quadratic Formula t. A quadratic equation written in standard form, ax 2 + bx + c = 0, has the solutions. Martin-Gay, Developmental Mathematics 10

Solving Quadratic Equations t Steps in Solving Quadratic Equations 1) If the equation is in the form (ax+b)2 = c, use the square root property to solve. 2) If not solved in step 1, write the equation in standard form. 3) Try to solve by factoring. 4) If you haven’t solved it yet, use the quadratic formula. Martin-Gay, Developmental Mathematics 11

The Quadratic Formula Example t. Solve 11 n 2 – 9 n = 1 by the quadratic formula. t 11 n 2 – 9 n – 1 = 0, so t a = 11, b = -9, c = -1 Martin-Gay, Developmental Mathematics 12

The Quadratic Formula Example Solve x 2 + x – = 0 by the quadratic formula. x 2 + 8 x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = -20 Martin-Gay, Developmental Mathematics 13

Solving Equations Example t. Solve the following quadratic equation. Martin-Gay, Developmental Mathematics 14

The Quadratic Formula Example x(x + 6) = -30 by the quadratic formula. t x 2 + 6 x + 30 = 0 t a = 1, b = 6, c = 30 t. Solve So there is no real solution. Martin-Gay, Developmental Mathematics 15

The Discriminant t. The expression under the radical sign in the formula (b 2 – 4 ac) is called the discriminant. t. The discriminant will take on a value that is positive, 0, or negative. t. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively. Martin-Gay, Developmental Mathematics 16

The Discriminant Example t. Use the discriminant to determine the number and type of solutions for the following equation. t 5 – 4 x + 12 x 2 = 0 t a = 12, b = – 4, and c = 5 tb 2 – 4 ac = (– 4)2 – 4(12)(5) t = 16 – 240 t = – 224 t. There are no real solutions. Martin-Gay, Developmental Mathematics 17

Solving Equations Example 3 x = x 2 + 1. t 0 = x 2 – 3 x + 1 t. Let a = 1, b = -3, c = 1 t. Solve Martin-Gay, Developmental Mathematics 18