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Quadratic Equation

Quadratic Equation. A quadratic equation is an equation equivalent to one of the form Where a , b , and c are real numbers and a  0. So if we have an equation in x and the highest power is 2, it is quadratic.

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Quadratic Equation

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  1. Quadratic Equation

  2. A quadratic equation is an equation equivalent to one of the form Where a, b, and c are real numbers and a 0 So if we have an equation in x and the highest power is 2, it is quadratic. To solve a quadratic equation we get it in the form above and see if it will factor. Get form above by subtracting 5x and adding 6 to both sides to get zero on right side. -5x + 6 -5x + 6 Factor of 6 {1,6 or 2,3} Use the Null Factor law and set eachfactor = 0 and solve.

  3. Remember standard form for a quadratic equation is: In this form we could have the case where b = 0. When this is the case, we get the x2 alone and then square root both sides. Getx2 alone by adding 6 to both sides and then dividing both sides by 2 + 6 + 6 Now take the square root of both sides remembering that you must consider both the positive and negative root. Now take the square root of both sides remembering that you must consider both the positive and negative root.  2 2 Let's check:

  4. We could factor by pulling an x out of each term. What if in standard form, c = 0? Factor out the common x Use the Null Factor law and set each factor = 0 and solve. If you put either of these values in for x in the original equation you can see it makes a true statement.

  5. Solving Quadratic Equations by Factoring Example

  6. Example

  7. Example

  8. Example

  9. Example

  10. Solving Quadratic Equations by square roots An alternate method of solving a quadratic equation is using the Principle of Taking the Square Root of Each Side of an Equation If x2 = a, then x = +

  11. Ex: Solve by taking square roots 3x2 – 36 = 0 3x2 – 36 = 0 First, isolate x2: 3x2 = 36 x2 = 12 Now take the square root of both sides:

  12. Ex: Solve by taking square roots 4(z – 3)2 = 100 First, isolate the squared factor: 4(z – 3)2 = 100 (z – 3)2 = 25 Now take the square root of both sides: z – 3 = + 5 z = 3 + 5  z = 3 + 5 = 8 and z = 3 – 5 = – 2

  13. Ex: Solve by taking square roots 5(x + 5)2 – 75 = 0 First, isolate the squared factor: 5(x + 5)2 = 75 (x + 5)2 = 15 Now take the square root of both sides:

  14. Steps to solving Quadratic Equations by Completing the Square • Write the equation in the form • Add to each side of the equation • Factor the perfect-square trinomial • Take the square root of both sides of the equation • Solve for x

  15. Ex: Solve x2+ 6x + 4 = 0 by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. x2+ 6x = – 4 6 Add to both sides:  = 32 = 9 b = 6 x2+ 6x + 9 = – 4 + 9 x2+ 6x + 9 = 5 (x + 3)2 = 5 Now take the square root of both sides

  16. solve by completing the square. To complete the square we want the coefficient of the x2 term to be 1. 2 2 2 2 Divide everything by 2 16 16 Since it doesn't factor get the constant on the other side ready to complete the square. So what do we add to both sides? the middle term's coefficient divided by 2 and squared Factor the left hand side Square root both sides (remember ) Add 4 to both sides to get x alone

  17. Ex: Solve 2y2= 3 – 5y by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. 2y2+ 5y = 3  y2+ (5/2)y = (3/2) (5/2) Add [½(b)]2 to both sides: b = 5/2 • [½(5/2)]2 = (5/4)2 = 25/16 y2+ (5/2)y + 25/16 = (3/2) + 25/16 y2+ (5/2)y + 25/16 = 24/16 + 25/16 (y + 5/4)2 = 49/16 Now take the square root of both sides

  18. y= - (5/4) + (7/4) = 2/4 = ½ and y= - (5/4) - (7/4) = -12/4 = - 3 y= { ½ , - 3}

  19. Ex: Solve 3n– 5 = (n – 1)(n – 2) Is this a quadratic equation? 3n – 5 = n2– 2n – n + 2 Collect all terms 3n – 5 = n2– 3n + 2 n2– 6n + 7 = 0 n2– 6n = – 7  [½(-6)]2 = (-3)2 = 9 n2– 6n + 9 = – 7 + 9 (n – 3)2 = 2

  20. Solving Quadratic Equations by Quadratic Formula • Consider a quadratic equation of the formax2 + bx + c = 0 for (a≠0). • Completing the square

  21. Solutions to ax2 + bx + c = 0 for (a≠0) are:

  22. WHY USE THE QUADRATIC FORMULA? The quadratic formula allows you to solve any quadratic equation, even if you cannot factor it. An important piece of the quadratic formula is what’s under the radical: b2 – 4ac This piece is called the discriminant.

  23. WHY IS THE DISCRIMINANT IMPORTANT? The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

  24. WHAT THE DISCRIMINANT TELLS YOU!

  25. Example Use thediscriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224<0 There are no real solutions.

  26. Example 6 – 3x2 –2x = 0 a = – 3, b = –2, and c = 6 b2 – 4ac = (–2)2 – 4(– 3)(6) = 4 + 72 = 76>0 There are 2 distinct root .

  27. Example x2 –4x +4= 0 a = 1, b = – 4, and c = 4 b2 – 4ac = (–4)2 – 4(1)(4) = 16 – 16 = 0 We have two double root

  28. By completing the square on a general quadratic equation in standard form we come up with what is called the quadratic formula. This formula can be used to solve any quadratic equation whether it factors or not. If it factors, it is generally easier to factor---but this formula would give you the solutions as well. We solved this by completing the square but let's solve it using the quadratic formula 1 6 6 (1) (3) (1) Don't make a mistake with order of operations! Let's do the power and the multiplying first.

  29. There's a 2 in common in the terms of the numerator These are the solutions we got when we completed the square on this problem. NOTE: When using this formula if you've simplified under the radical and end up with a negative, there are no real solutions.(There are complex (imaginary) solutions, but that will be dealt with in year 12 Calculus).

  30. ax2 + bx + c = 0 for (a≠0). =b2 – 4ac • If,then • If,then we have two complex number (no-Real –Root) • If,then

  31. If b is an even number in the quadratic Equation ax2+ bx + c = 0 for (a≠0). =–ac where = • If,then • If ,then we have two complex number (no-Real –Root) • If,then

  32. Sum and Product of the Root The sum of the roots of the Quadratic equation:ax2 + bx + c = 0is And the product of roots is This means that: P

  33. Note: Two numbers whose given sum is S and given product is P are roots of the equation x2 - Sx+ P= 0 Example: Find two numbers knowing their sum S and their product P : S=4 and P=4 The two required numbers are the solutions of the equation: x2- 4x + 4= 0,that is (x-2)2=0,which gives x’=x”=2

  34. Example Write the form of a quadratic equation whose roots are x’= and x”= Solution: The sum S=x’+x”= And the product P=x’.x”= The equation :

  35. Example Solve using the Quadratic Formula

  36. Ex: Use the Quadratic Formula to solve x2 + 7x + 6 = 0 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in ax2+ bx + c = 0: a = b = c = 1 7 6 Now evaluate the quadratic formula at the identified values of a, b, and c

  37. x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6 x = { - 1, - 6 }

  38. Ex: Use the Quadratic Formula to solve 2m2 + m – 10 = 0 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in am2 + bm + c = 0: a = b = c = 2 1 - 10 Now evaluate the quadratic formula at the identified values of a, b, and c

  39. m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2 m = { 2, - 5/2 }

  40. Example Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1

  41. Solve x2 + x – = 0 by the quadratic formula. Example x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20

  42. Example Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.

  43. Example Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = -3, c = 1

  44. Example Solve the following quadratic equation.

  45. Solving Quadratic Equations by the Quadratic Formula Try the following examples. Do your work on your paper and then check your answers.

  46. Quadratic Equations and Problem Solving A cliff diver is 64 feet above the surface of the water. The formula for calculating the height (h) of the diver after t seconds is: How long does it take for the diver to hit the surface of the water? seconds

  47. Quadratic Equations and Problem Solving The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. What are the length and the width of the garden? The width is w. The length is w+5. feet feet

  48. Quadratic Equations and Problem Solving Find two consecutive odd numbers whose product is 23 more than their sum? Consecutive odd numbers:

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