Chapter 2: Equations and Inequalities 2.2: Solving Equations Algebraically

1. Chapter 2: Equations and Inequalities2.2: Solving Equations Algebraically Essential Question: What are some things the discriminate is used for?

2. 2.2 Solving Equations Algebraically • 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: • ax2 + bx + c = 0 • For real constants a, b, and c, with a ≠ 0

3. 2.2 Solving Equations Algebraically • 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: • Factoring • Taking the square root of both sides of an equation • Techniques that can be used to solve all quadratic equations • Completing the square • Using the quadratic formula

4. 2.2 Solving Equations Algebraically • Factoring • Rearrange the terms so that everything equals 0 • Find two numbers that multiply together to get a•c & add together to get b • Use those numbers to split b (the term in the middle) • Take out the greatest common factor in each group • Group the outside terms together • Set each part equal to 0 and solve • Example 1: Solve 3x2 – x = 10 by factoring • 3x2 – x – 10 = 0 [ax2 + bx + c = 0] • Numbers that multiply to get -30, add to get -1?-6 and 5 • (3x2– 6x) + (5x – 10) = 0 • 3x(x – 2) + 5(x – 2) = 0 • (3x + 5)(x – 2) = 0 • x = -5/3 or x = 2

5. 2.2 Solving Equations Algebraically • Taking the square root of both sides • Only works when x2 = k(a positive constant - no “bx” term) • For a real number k:

6. 2.2 Solving Equations Algebraically • Taking the square root of both sides • Example 2: Solve 3x2 = 9 • Get squared term by itself • x2 = 3 • Take the square root of both sides • x =

7. 2.2 Solving Equations Algebraically • Taking the square root of both sides • Example 3: Solve 2(x + 4)2 = 6 • Get squared term by itself • (x + 4)2 = 3 • Take the square root of both sides • x + 4 = • Get x by itself • x = -4

8. 2.2 Solving Equations Algebraically • Assignment • Page 95, 1-23 (odds) • Don’t expect to get credit without showing work

9. Chapter 2: Equations and Inequalities2.2: Solving Equations AlgebraicallyPart 2 Essential Question: What are some things the discriminate is used for?

10. 2.2 Solving Equations Algebraically • Note: Completing the square is really only useful for determining the quadratic equation. We’ll do that after this short demonstration… • Completing the square • Write the equation in the form x2 + bx = c • Add to both sides so that the left side is a perfect square and the right side is a constant • Take the square root of both sides • Simplify

11. 2.2 Solving Equations Algebraically • Completing the square • Example 4: Solve 2x2 – 6x + 1 = 0 by completing the square • Write the equation in the form x2 + bx = c • 2x2 – 6x = -1 • x2 – 3x = -½ • Add to both sides so that the left side is a perfect square and the right side is a constant

12. 2.2 Solving Equations Algebraically • Completing the square (continued) • Take the square root of both sides • Simplify

13. 2.2 Solving Equations Algebraically • The Quadratic Formula • The solutions of the quadratic equationax2 + bx + c = 0 are • Get an equation to equal 0, then simply substitute in the formula

14. 2.2 Solving Equations Algebraically • Solve x2 + 3 = -8x by using the quadratic formula • Get equation to equal 0 • x2 + 8x + 3 = 0 • Plug into the quadratic formula • a = 1, b = 8, c = 3

15. 2.2 Solving Equations Algebraically • The Discriminant • The portion of the quadratic formula that exists underneath the square root (b2 – 4ac) is called the discriminant. It can be used to determine the number of real solutions of a quadratic equation.

16. 2.2 Solving Equations Algebraically • Using the discriminant • Find the number of real solutions to 2x2 = -x – 3 • Write the equation in general form(make one side = 0) • 2x2 + x + 3 = 0 • Plug into the discriminant and simplify • a = 2, b = 1, c = 3 • b2 – 4ac • (1)2 – 4(2)(3) = 1 – 24 = -23 • Because -23 < 0, the equation has no real solutions • We’ll confirm using the calculator

17. 2.2 Solving Equations Algebraically • Polynomial Equations • A polynomial equation of degree n is an equation that can be written in the form: • anxn + an-1xn-1 + … + + a1x + a0 = 0 • Example: 4x6 – 3x5 + x4 + 7x3 – 8x2 + 4x + 9 = 0 is a polynomial equation of degree 6. • Example 2: 4x3 – 3x2 + 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.

18. 2.2 Solving Equations Algebraically • Polynomial Equations in Quadratic Form • Solve 4x4 – 13x2 + 3 = 0 • To solve, we substitute u for x2 • 4u2 – 13u + 3 = 0 • Then solve the quadratic equation • (4u – 1)(u – 3) = 0 • u = ¼ or u = 3 • Because u = x2:

19. 2.2 Solving Equations Algebraically • Assignment • Page 95-96, 25-53 (odds) • For the problems that direct you to solve by completing the square, use the quadratic formula instead. • Hints for 47-53 • 47, 49 & 51 can be factored (though you don’t have to solve by factoring) • Don’t expect to get credit without showing work