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Math 152B – Cass Gustafson – 8.3 The Discriminant and Equations That Can Be Written in Quadratic Form I. The Discriminant A. The radicand b2 – 4ac in the quadratic formula is called the discriminant. B. It is called the discriminant because: Discriminant b2 – 4ac Solutions to ax2 + bx + c = 0 2 The equation has two unequal REAL solutions. 2 If b – 4ac is a perfect square, the solutions are RATIONAL numbers. 2 If b – 4ac is NOT a perfect square, the solutions are IRRATIONAL CONJUGATES. 2 The equation has two equal REAL solutions, called a double root and if a, b, and c are rational, it would be a RATIONAL number. 2 The equation has NO REAL solutions. It has two IMAGINARY solutions. They would be COMPLEX CONJUGATES. If b – 4ac > 0 If b – 4ac = 0 If b – 4ac < 0 Ex. 1: For each equation, compute the discriminant. Then determine the number and type of solutions. Do not solve the equations. a) 2x2 – 7x – 4 = 0 b) 3x2 – 2x + 4 = 0 Ex. 2: Find the value of k that will make the solutions of kx2 – 20x + 25 = 0 equal. II. Determining Which Method to Use: What method do we use to solve a quadratic equation? Form of the Quadratic Equation Most Efficient Solution Method ax2 + bx + c = 0 and ax2 + bx + c can be factored easily Solve by factoring. ax2 + c = 0 (The equation has no “x” term, b = 0) Solve using the square root property (a polynomial)2 = a number Solve using the square root property 2 Solve by completing the square or by using the quadratic formula. 2 ax + bx = c = 0 and ax + bx + c cannot be factored or is too difficult to factor x = −b ± b 2 − 4ac 2a III. Solving Equations That Can Be Written in Quadratic Form When equations are quadratic in form, they will be of the pattern au2 + bu + c = 0 where u can be any algebraic expression. Some examples follow: Example (x 2 Quadratic-Like x 4 − 9x 2 + 8 = 0 (x ) x − 3 x − 4=0 ( x) ) −1 2 ( ) − x2 − 1 − 2 = 0 m−2 − 6m−1 + 4 = 0 2 n5 − n 1 5 − 2=0 2 (x 2 2 2 ) −1 (m ) −1 1 5 − 3 2 2 ( ) n ( ) − 9 x2 In General u2 – 9u + 8 = 0 + 8=0 ( x) − ( ) u2 – u – 2 = 0 − x2 − 1 − 2 = 0 ( ) − 6 m−1 + 4 = 0 2 u2 – 3u – 4 = 0 4=0 ( ) 1 − n5 u2 – 6u + 4 = 0 u2 – u – 2 = 0 − 2=0 What is “u” in each of the examples above? Ex. 3: Solve the following equations. Notice that sometimes we must check our solutions! a) x + Gustafson – 8.3 x − 12 = 0 b) x 2 3 = − 3x 1 3 + 10 2 c) 12 6 + = 5 x x+3 d) 28x −2 − 3x −1 − 1 = 0 IV. Solving Equations for a Specified Variable Ex. 4: Solve each equation for the indicated variable. a) a2 + b2 = c2 for a b) kx = ay – x2 for x V. Verifying Solutions of a Quadratic Equation by Calculating the Sum and Product This subsection is OPTIONAL. Solve problems 63-69 (odd), but don’t do the verification. Consider the solutions of a quadratic equation based on the quadratic formula … r1 = −b + b2 − 4ac 2a r2 = −b − b2 − 4ac 2a Calculate the sum … Calculate the product … Gustafson – 8.3 3 We can use this to state that if r1 + r2 = − b a and r1 r2 = c a then r1 and r2 are solutions of ax2 + bx + c = 0. We can also use these formulas to check our answers. Ex. 5: Solve 8x2 – 2x – 3 = 0 and verify that the sum of the solutions is − the solutions is Gustafson – 8.3 b and that the product of a c . a 4