1 / 37

Equations: Linear and Quadratic Solutions

Learn how to solve linear and quadratic equations. Understand the difference between linear and nonlinear equations. Use factoring, completing the square, and the quadratic formula to find the solutions. Practice with examples and check your answers.

evajohnson
Download Presentation

Equations: Linear and Quadratic Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Objectives • Solving Linear Equations • Solving Quadratic Equations • Other Types of Equations

  2. Equations • The values of the unknown that make the equation true are called the solutions or roots of the equation, and the process of finding the solutions is called solving the equation. • Two equations with exactly the same solutions are called equivalent equations. To solve an equation, we try to find a simpler, equivalent equation in which the variable stands alone on one side of the “equal” sign.

  3. Equations

  4. Solving Linear Equations • the difference between linear and nonlinear equations.

  5. Example 1 – Solving a Linear Equation • Solve the equation 7x – 4 = 3x + 8. • Solution:We solve this by changing it to an equivalent equation with all terms that have the variable x on one side and all constant terms on the other. • 7x – 4 = 3x + 8 • (7x – 4) + 4= (3x + 8) + 4 Given equation Add 4

  6. Example 1 – Solution cont’d • 7x = 3x + 12 • 7x – 3x = (3x + 12) – 3x • 4x = 12 • x = 3 Simplify Subtract 3x Simplify Multiply by Simplify

  7. Example 1 – Solution cont’d • Check Your Answer: • x = 3: LHS = 7(3) – 4 RHS = 3(3) + 8 • = 17 = 17 • LHS = RHS

  8. Solving Quadratic Equations

  9. Solving Quadratic Equations • Linear equations are first-degree equations like 2x + 1 = 5 or 4 – 3x = 2. Quadratic equations are second-degree equations like x2 + 2x – 3 = 0 or 2x2 + 3 = 5x.

  10. Solving Quadratic Equations • Set equation = to 0 Then Factor & Use….. • This means that if we can factor the left-hand side of a, • Then we can solve it by setting each factor equal to 0 in turn. • This method worksonly when the right-hand side of the equation is 0.

  11. Example 4 – Solving a Quadratic Equation by Factoring • Solve the equation x2 + 5x = 24. • Solution: rewrite the equation so the right-hand side is 0. • x2 + 5x – 24 = 0 • (x – 3)(x + 8) = 0 • x – 3 = 0 or x + 8 = 0 Zero-Product Property • x = 3 x = –8 Solve • And Check your answers!!! Subtract 24 Factor

  12. Example 5 – Solving Simple Quadratics • (a) x2 = 5 (b) (x – 4)2 = 5 • Solution:(a) • (b)Take the square root of each side of the equation. • (x – 4)2 = 5 • Take the square root

  13. Solving Quadratic Equations • if a quadratic equation is of the form (xa)2 = c we can take the square root of each side…. • If a quad equation does not factor , complete the square. • Add a constant to make it a perfect square. • To make x2 – 6x a perfect square, • add 9, since x2 – 6x + 9 = (x – 3)2.

  14. Example 6 – Solving Quadratic Equations by Completing the Square • Solve each equation. • (a) x2 – 8x + 13 = 0 (b) 3x2 – 12x + 6 = 0 • Solution: • (a) x2 – 8x + 13 = 0 • x2 – 8x = –13 • x2 – 8x + 16 = –13 + 16 • Perfect square • Take square root • Add 4 Given equation Subtract 13 Complete the square: add = 16

  15. Example 6 – Solution cont’d • 3x2 – 12x + 6 = 0 • 3x2 – 12x = –6 • 3(x2 – 4x)= –6 • we complete the square by adding (–2)2 = 4 inside the parentheses. Since everything inside the parentheses is multiplied by 3, this means that we are actually adding 3  4 = 12 to the left side of the equation. Given equation Subtract 6 Factor 3 from LHS

  16. Example 6 – Solution cont’d • Thus, we must add 12 to the right side as well. • 3(x2 – 4x + 4) = –6 + 3 4 • 3(x– 2)2 = 6 • (x– 2)2 = 2 Complete the square: add 4 Perfect square Divide by 3 Take square root Add 2

  17. Solving Quadratic Equations • We can use the technique of completing the square to derive a formula for the roots of the general quadratic equation ax2 + bx + c = 0.

  18. Example 7 – Using the Quadratic Formula • Find all solutions of each equation. • (a) 3x2 – 5x – 1 = 0 (b) 4x2 + 12x + 9 = 0 • (c) x2 + 2x + 2 = 0 • Solution:(a) In this quadratic equation a = 3, b =–5, and c =–1. • 3x2 – 5x – 1 = 0

  19. Example 7 – Solution cont’d • By the Quadratic Formula, • If approximations are desired, we can use a calculator to obtain • and

  20. Example 7 – Solution cont’d • (b) Using the Quadratic Formula with a = 4, b = 12, and c = 9 gives • This equation has only one solution,

  21. Example 7 – Solution cont’d • (c) Using the Quadratic Formula with a = 1, b = 2, and c = 2 gives

  22. Solving Quadratic Equations

  23. Example 8 – Using the Discriminant • Use the discriminant to determine how many real solution each equation has. • (a) x2 + 4x – 1 = 0 (b) 4x2 – 12x + 9 = 0 • (c) x2–2x + 4 = 0 • Solution: • (a) The discriminant is D = 42 – 4(1)(–1) = 20 > 0, so the equation has two distinct real solutions. • (b) The discriminant is D = (–12)2 – 4  4  9 = 0, so the equation has exactly one real solution.

  24. Example 8 – Solution cont’d • (c) The discriminant is D = (–2)2 – so the equation has no real solution.

  25. Example 10 – An Equation Involving Fractional Expressions • Solve the equation • Solution:We eliminate the denominators by multiplying each side by the lowest common denominator. • x(x + 2) = 2x(x + 2) • 3(x + 2) + 5x = 2x2 + 4x Multiply by LCD x(x + 2) Expand

  26. Example 10 – Solution cont’d • 8x + 6 = 2x2 + 4x • 0 = 2x2 – 4x – 6 • 0 = x2 – 2x – 3 • 0 = (x – 3)(x + 1) • x – 3 = 0 or x + 1 = 0 • x = 3 x = –1 Expand LHS Subtract 8x+ 6 Divide both sides by 2 Factor Zero-Product Property Solve

  27. Example 10 – Solution cont’d • We must check our answers because multiplying by an expression that contains the variable can introduce extraneous solutions. From Check Your Answers we see that the solutions are x = 3 and –1. • Check Your Answers: • x = 3: • LHS = = 1 + 1 = 2 RHS = 2 • LHS = RHS

  28. Example 10 – Solution cont’d • x = –1: • LHS = = –3 + 5 = 2 • RHS = 2 • LHS = RHS

  29. Example 11 – An Equation Involving a Radical • Solve the equation • Solution:To eliminate the square root, we first isolate it on one side of the equal sign, then square: • (2x – 1)2 = 2 – x • 4x2 – 4x + 1 = 2 – x Subtract 1 Square each side Expand LHS

  30. Example 11 – Solution cont’d • 4x2 – 3x – 1 = 0 • (4x + 1)(x – 1) = 0 • 4x + 1 = 0 or x – 1 = 0 • x = x = 1 • The values x = and x = 1 are only potential solutions. We must check them to see if they satisfy the original equation. From Check Your Answers we see that x = is a solution but x = 1 is not. The only solution is x = . Add –2 + x Factor Zero-Product Property Solve

  31. Example 11 – Solution cont’d • Check Your Answers: • LHS = • RHS = • LHS = RHS

  32. Example 11 – Solution cont’d • x = 1: • LHS = 2(1) • = 2 • RHS = • = 1 – 1 • = 0 • LHS ≠ RHS

  33. Other Types of Equations • When we solve an equation, we may end up with one or more extraneous solutions, that is, potential solutions that do not satisfy the original equation. • In Example 11 the value x = 1 is an extraneous solution. Extraneous solutions may be introduced when we square each side of an equation because the operation of squaring can turn a false equation into a true one. • For example, –1 ≠ 1, but (–1)2 = 12. Thus, the squared equation may be true for more values of the variable than the original equation.

  34. Other Types of Equations • That is why you must always check your answers to make sure that each satisfies the original equation. • An equation of the form aW2 + bW + c = 0, where W is an algebraic expression, is an equation of quadratic type. We solve equations of quadratic type by substituting for the algebraic expression, as we see in the next example.

  35. Example 12 – A Fourth-Degree Equation of Quadratic Type • Find all solutions of the equation x4 – 8x2 + 8 = 0. • Solution:If we set W = x2, then we get a quadratic equation in the new variable W: • (x2)2 – 8x2 + 8 = 0 • W2 – 8W + 8 = 0 Write x4 as (x2)2 Let W = x2

  36. Example 12 – Solution cont’d Quadratic Formula W = x2 Take square roots

  37. Example 12 – Solution cont’d • So, there are four solutions: • Using a calculator, we obtain the approximations x  2.61, 1.08, –2.61, –1.08.

More Related