Essential Question: What are some things the discriminate is used for?

 Basic strategy ◦ Add or subtract the same quantity from both sides of the equation ◦ Multiply or divide both sides of the equation by the same nonzero quantity.  Definition of a Quadratic Equation ◦ A quadratic, or second degree, equation is one that can be written in the form:  ax 2 + bx + c = 0 ◦ For real constants a, b, and c, with a ≠ 0

 Techniques used to solve quadratic equations ◦ There are four techniques used to algebraically find exact solutions of quadratic equations. ◦ Techniques that can be used to solve some quadratic equations: 1.Factoring 2.Taking the square root of both sides of an equation ◦ Techniques that can be used to solve all quadratic equations 3.Completing the square 4.Using the quadratic formula

 Factoring 1.Rearrange the terms so that everything equals 0 2.Find two numbers that multiply together to get ac & add together to get b 3.Use those numbers to split b (the term in the middle) 4.Take out the greatest common factor in each group 5.Group the outside terms together 6.Set each part equal to 0 and solve  Example 1: Solve 3x 2 – x = 10 by factoring 1)3x 2 – x – 10 = 0 [ax 2 + bx + c = 0] 2)Numbers that multiply to get -30, add to get -1? -6 and 5 3)(3x 2 – 6x) + (5x – 10) = 0 4)3x(x – 2) + 5(x – 2) = 0 5)(3x + 5)(x – 2) = 0 6)x = -5/3 or x = 2

 Taking the square root of both sides ◦ Only works when x 2 = k (a positive constant - no “bx” term) ◦ For a real number k: Number of SolutionsSolutions k < 00 k = 010 k > 02

 Taking the square root of both sides ◦ Example 2: Solve 3x 2 = 9 1.Get squared term by itself  x 2 = 3 2.Take the square root of both sides  x =

 Taking the square root of both sides ◦ Example 3: Solve 2(x + 4) 2 = 6 1.Get squared term by itself  (x + 4) 2 = 3 2.Take the square root of both sides  x + 4 = 3.Get x by itself  x = -4

 Assignment ◦ Page 95, 1-23 (odds)  Don’t expect to get credit without showing work

Essential Question: What are some things the discriminate is used for?

 Note: Completing the square is really only useful for determining the quadratic equation. We’ll do that after this short demonstration…  Completing the square 1.Write the equation in the form x 2 + bx = c 2.Add to both sides so that the left side is a perfect square and the right side is a constant 3.Take the square root of both sides 4.Simplify

 Completing the square ◦ Example 4: Solve 2x 2 – 6x + 1 = 0 by completing the square 1.Write the equation in the form x 2 + bx = c  2x 2 – 6x = -1  x 2 – 3x = -½ 2.Add to both sides so that the left side is a perfect square and the right side is a constant 

 Completing the square (continued) 1.Take the square root of both sides  2.Simplify 

 The Quadratic Formula ◦ The solutions of the quadratic equation ax 2 + bx + c = 0 are ◦ Get an equation to equal 0, then simply substitute in the formula

 Solve x = -8x by using the quadratic formula ◦ Get equation to equal 0  x 2 + 8x + 3 = 0 ◦ Plug into the quadratic formula  a = 1, b = 8, c = 3 

 The Discriminant ◦ The portion of the quadratic formula that exists underneath the square root (b 2 – 4ac) is called the discriminant. It can be used to determine the number of real solutions of a quadratic equation. Discriminant ValueNumber of Real Solutions of ax 2 + bx + c = 0 b 2 – 4ac < 00 real solutions b 2 – 4ac = 01 distinct real solution b 2 – 4ac > 02 distinct real solutions

 Using the discriminant ◦ Find the number of real solutions to 2x 2 = -x – 3 1.Write the equation in general form (make one side = 0)  2x 2 + x + 3 = 0 2.Plug into the discriminant and simplify  a = 2, b = 1, c = 3  b 2 – 4ac  (1) 2 – 4(2)(3) = 1 – 24 = -23 ◦ Because -23 < 0, the equation has no real solutions ◦ We’ll confirm using the calculator

 Polynomial Equations ◦ A polynomial equation of degree n is an equation that can be written in the form:  a n x n + a n-1 x n-1 + … + + a 1 x + a 0 = 0 ◦ Example: 4x 6 – 3x 5 + x 4 + 7x 3 – 8x 2 + 4x + 9 = 0 is a polynomial equation of degree 6. ◦ Example 2: 4x 3 – 3x 2 + 4x - 5 = 0 is a polynomial expression of degree 3. ◦ Polynomials have the following traits  No variables in denominators (integers only)  No variables under radical signs ◦ Polynomials of degree 3 and above are best solved graphically. However, some equations are quadratic in form and can be solved algebraically.

 Polynomial Equations in Quadratic Form ◦ Solve 4x 4 – 13x = 0 ◦ To solve, we substitute u for x 2  4u 2 – 13u + 3 = 0 ◦ Then solve the quadratic equation  (4u – 1)(u – 3) = 0  u = ¼ or u = 3 ◦ Because u = x 2 : 

 Assignment ◦ Page 95-96, (odds)  For the problems that direct you to solve by completing the square, use the quadratic formula instead.  Hints for  47, 49 & 51 can be factored (though you don’t have to solve by factoring)  Don’t expect to get credit without showing work