1	Quadratic Equations
2	Quadratic Equations A quadratic equation is of the form axⁿ + bx + c = 0, where a, b, and c are real numbers and n > 1. Quadratic equations can be solved by Factoring Completing the square Using the quadratic formula
3	Factor to Solve for x Factor ax² + bx + c = 0 by trial and error a, b, and c are real numbers Set both factors equal to zero to solve for x Example: 2x² + 3x - 2 = 0 factors as (2x - 1)(x + 2) = 0 The equation is true when either factor equals zero 2x - 1 = 0 x + 2 = 0 2x = 1 x = - 2
4	Completing the Square Rewrite ax² + bx + c = 0 in the form a(x - h)² + k = 0 Example: 2x² + 3x - 2 = 0 Group the first and second degree terms together (2x² + 3x) - 2 = 0 Factor out the coefficient of the second degree term (a) 2(x² + x) - 2 = 0 Add the constant, c, to both sides 2(x² + x) = 2 (continued on next slide)
5	Completing the Square page 2 2(x² + x) = 2 Take half the resulting coefficient of the first degree term ( ) ( ) · ( ) = Square the result ( )² = Add in and subtract back out this new amount inside the parentheses 2(x² + x + - ) = 2 (continued on next slide)
6	Completing the Square page 3 Determine the value of the last term inside the parentheses 2(- ) = - (multiply by the outside factor) Remove this last term from inside the parentheses 2(x² + x + ) - = 2 Bring this value to the other side of the equation and reduce 2(x² + x + ) = 2 + = If you have trouble with fractions or reducing, review those lessons. (continued on next slide)
7	Completing the Square page 4 2(x² + x + ) = Express the quantity inside the parentheses as the square of a binomial 2(x + )² = (trial and error factoring) Divide both sides of the equation by the coefficient of the squared term (x + )² = = Take the square root of both sides of the equation x + = ± (continued on next slide)
8	Completing the Square page 5
9	Quadratic Formula to Solve for x The equation ax² + bx + c = 0 can be solved by substituting values for a, b, and c into the formula x = -b ± b² - 4ac 2a This formula is known as the quadratic formula. The quantity under the radical sign is known as the discriminant, and helps determine the nature of the resulting values for x.
10	Information from the Discriminant Given an equation of the form ax² + bx + c = 0 where b² - 4ac (the discriminant) has been calculated, the nature of the solutions are as follows: Value of Discriminant Nature of Solutions positive 2 real zero 1 real negative 2 non-real
11	Complex numbers Quadratic Equations may have non-real complex solutions a + bi represents a complex number, where a and b are real numbers If b = 0, the complex number is a real number If b ¹ 0, the complex number is then a non-real number i, by definition, is the square root of -1. It is an imaginary number.
12	"Example:" Example: x² - 6x + 13 = 0 This equation is not easily factored. It can be solved by completing the square or by using the quadratic formula. By using the quadratic formula, the solutions are: a = 1, b = -6, c = 13
13	Solving Quadratics: Exercises Solve -4x²+ x + 14 = 0 by factoring ( 4x + 7) (-x + 2) = 0 answers: x = , x = 2 Solve 4x² - 4x + 1 = 9 by completing the square 4(x² - x + 1/4 ) = 9 (lots of steps!) (x - 1/2 )²= 9/4 x - 1/2 = + 3/2 x = 2 and x = -1
14	Solving Quadratics: Exercises page 2 Solve 7x² + 5 = 0 by the quadratic formula x = -b ± Ö b² - 4ac 2a x = -0 ± Ö 0²- 4 7 5 a = 7, b = 0, c = 5 14 x = ± Ö -140 14 x = ± 2i Ö 35 = + i Ö 35 14 7