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0012 chapter v

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0012 chapter v

  1. 1. Chapter V<br />The discriminant gives additional information on the nature of the roots beyond simply whether there are any repeated roots: it also gives information on whether the roots are real or complex, and rational or irrational. More formally, it gives information on whether the roots are in the field over which the polynomial is defined, or are in an extension field, and hence whether the polynomial factors over the field of coefficients. This is most transparent and easily stated for quadratic and cubic polynomials; for polynomials of degree 4 or higher this is more difficult to state.<br />TARGET SKILLS:<br />At the end of this chapter, students are expected to:<br />• determine discriminant, roots and coefficient;<br />• discuss the relation the roots and coefficient;<br />• find the sum and product of the roots; and<br />• change quadratic equation to discriminant formula.<br />Lesson 13<br />The Discriminant and the roots of a<br />Quadratic Equation<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>determine discriminant and the roots;
  2. 2. compare discriminant and the nature of the roots; and
  3. 3. change quadratic equation to discriminant using the nature of the roots.</li></ul>Example<br /> 1. Find the x-intercept of y = 3x² - 6x + 4.<br />Solution: As already mentioned, the values of x for which 3x² - 6x + 4 = 0 give the x-intercepts of the function. We apply the quadratic formula in solving the equation.<br />3x² - 6x + 4 = 0<br />x = -(ˉ6)±(ˉ6)-4(3)(4)2(3) = 6±ˉ126<br />Since ˉ12 is not a real number, the equation 3x² - 6x + 4 = 0 has no real root. This means that the parabola y = 3x² - 6x + 4 does not intersect the x-axis.<br />Let us write the equation in the form y = a(x – h)²+ k.<br />y = 3(x² – 2x)² + 4<br /> = 3(x – 1)² + 1<br /> ∆ = b² - 4ac Roots of ax² + bx + c = 0PositiveReal and distinct r = -b-∆2a s = -b+∆2aZeroReal and equal r = s = ˉb2aNegativeNo real roots<br />Example 2. Use the disciminant to determine the nature of the roots of the <br /> following quadratic equation.<br />x² - x + ¼ = 0<br />a = 1, b = ˉ1, c = ¼ <br /> b² - 4ac = (ˉ1)² - 4 (1)(¼)<br /> = 1 – 1<br /> = 0<br />There is only one solution, that is, a double root.<br />Note that x² - x = ½ = (x - ½), so that double root is ˉb/2a = ½. <br />5x² - 4x + 1 = 0<br />a = 5, b = ˉ4, c = 1<br />b² - 4ac = (ˉ4)² - 4 (5)(1)<br /> = 16 – 20<br /> = ˉ4 < 0<br />There are no real roots since a negative number has no real square root.<br />Exercises:<br /> Solve each by using the discriminant.<br />x² + 3x - ½ = 0<br />3x² - 5x – 7 = 0<br />6x² + 3x + 8 = 0<br />x² + 9x – 6 = 0<br />8x² - 12x + 4 = 0<br />-406401-414118Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Use the Discriminant to determine the nature of the root of the following Quadratic Equations.<br /><ul><li>x2- 2x – 3=0
  4. 4. _____________________________________________________
  5. 5. 6x2 – x – 1 = 0
  6. 6. _____________________________________________________
  7. 7. 2x2 – 50 = 0
  8. 8. _____________________________________________________
  9. 9. x2 – 8x + 12 = 0
  10. 10. _____________________________________________________
  11. 11. x2 + 5x – 14 = 0
  12. 12. _____________________________________________________
  13. 13. -4x2 – 4x + 1 = 0
  14. 14. _____________________________________________________
  15. 15. 7x2 + 2x – 1 = 0
  16. 16. -440080-360539 _____________________________________________________
  17. 17. x2 + 3x = 40
  18. 18. _____________________________________________________
  19. 19. 3x2= 5x – 1
  20. 20. _____________________________________________________</li></ul>10. 3x2+ 12 – 1=0<br /><ul><li> _____________________________________________________</li></ul> 11. (x-2)(x-3) = 4<br /><ul><li> _____________________________________________________</li></ul>12.2x2 + 2x + 1 = 0<br /><ul><li> _____________________________________________________</li></ul>13. 7x2 + 3 – 6x = 0<br /><ul><li> _____________________________________________________</li></ul>14. 5x2 – 6x + 4 = 0<br /><ul><li> _____________________________________________________</li></ul>15. 3x2 + 2x + 2 = 0<br /><ul><li> _____________________________________________________</li></ul>Lesson 14 <br />Relation between roots and coefficient<br />OBJECTIVES:<br />At the end of this lesson, students are expected to:<br /><ul><li>classify roots and coefficient;
  21. 21. discuss relations between the roots and coefficient of the quadratic equation; and
  22. 22. find the sum and product of the roots of a given quadratic equation.</li></ul>There are some interesting relations between the sum and the product of the roots of a quadratic equation. To discover these, consider the quadratic equation ax2 + bx + c = 0, where a ≠ 0.<br />Multiply both sides of this equation by 1/a so that the coefficient of x2 is 1.<br />1a (ax2 + bx + c) = 1(0)a<br />We obtain an equivalent quadratic equation in the form <br />x2 + bxa + ca = 0<br />If r and s are the roots of the quadratic equation ax2 + bx + c = 0, then from the quadratic formula<br />r = -b+b2-4ac2a and s = -b-b2-4ac2a<br />Adding the roots, we obtain<br />r + s = -b+b2-4ac2a + -b-b2-4ac2a<br />= ˉ2b2a = ˉba<br />Multiplying the roots, we obtain<br />rs = (-b+b2-4ac)2a (-b-b2-4ac)2a<br />-b-(b² - 4ac)4a² = ca<br />Observe the coefficient in the quadratic equation x2 + bx/a + c/a = 0. How do they compare with the sum and the product of the roots? Did you observe the following?<br />The sum of the roots is equal to the negative of the coefficient of x.<br />r + s = ˉba<br />The product of the roots is equal to the constant term<br />rs = ca<br />An alternate way of arriving at these relations is as follows<br />Let r and s be the roots of x2 + bxa + ca = 0. Then <br />(x - r)(x – s) = 0<br />Expanding gives, x2 – rx – sx + rs = 0<br />or x2 – (r + s)x + rs = 0<br />Comparing the coefficients of the corresponding terms, we obtain<br />r + s = ˉba and rs = ca<br /> The above relations between the roots and the coefficients provide a fast and convenient means of checking the solutions of a quadratic equation.<br />Example: Solve and check. 2x2 + x – 6 = 0<br />Solutions: 2x2 + x – 6 = (2x – 3)(x + 2) = 0<br /> x = 32 or x = -2<br />The roots are 32 and -2.<br />To check, we add the roots, 32 = (-2) = ˉ12 = ˉba<br />and multiply them. 32 (-2) = -3 = ca<br />Example: Find the sum and the product of the roots of 3x2 – 6x + 8 = 0 without having to first determine the roots. <br />Solution: The sum of the roots is r + s = ˉba = ˉ(ˉ6)3 = 2;<br />and their product is rs = ca = 83.<br />Exercises: <br />Without solving for the roots, find the sum and product of the roots of the following:<br />x2 – 5x – 36 = 0<br />3x2 + 2x – 1 = 0<br />5x2 + x – 1 =0<br />2x+ x=3<br />xx-2=2x+1x+1<br />3px2 – 4pqx + x2 = 0 <br />-391885-561884Name: ___________________ Section: _______<br />Instructor: ________________ Date: _______ Rating: ____<br />Instruction: Without solving the roots, find the sum and product of the roots of the following.<br />6x2 – 5x + 2 = 0<br /><ul><li>_____________________________________________________</li></ul>x2 + x – 182 = 0<br /><ul><li>_____________________________________________________</li></ul>x2 – 5x – 14 = 0<br /><ul><li>_____________________________________________________</li></ul>2x2 – 9x + 8 = 0<br /><ul><li>_____________________________________________________</li></ul>3x2 - 5x – 2 = 0<br /><ul><li>_____________________________________________________</li></ul>X2 – 8x – 9 = 0<br /><ul><li>_____________________________________________________</li></ul>-429080-4487032x2 – 3x – 9 = 0<br /><ul><li>_____________________________________________________</li></ul>X2 + x – 2 - 2=0<br /><ul><li>_____________________________________________________</li></ul>3x2 + 2x – 8 = 0<br /><ul><li>_____________________________________________________</li></ul>16x2 – 24x + 12=0<br /><ul><li>_____________________________________________________</li></ul>x2 – 6x + 25 = 0<br /><ul><li>_____________________________________________________</li></ul>3x2 + x – 2 = 0 <br /><ul><li>_____________________________________________________</li></ul>5x2 + 11x – 8 = 0<br /><ul><li>_____________________________________________________</li></ul>x2 – 8x + 16 = 0<br /><ul><li>_____________________________________________________</li></ul>4x2 – 16x + 10 = 0<br /><ul><li>_____________________________________________________</li></ul>Use the discriminant to determine which of the following quadratic equations have two, one or no real roots. Give reasons.<br />x2 – 5x – 5 = 0<br />x2 – 3x – 2 = 3x – 11<br />5x2 – 9x = 2x – 7<br />2x2 – 7x + 8 = 0<br />3 – 4x – 2x2 = 0<br />4x2 – 9x + 5 = 3x – 7<br />By using the relations between roots and coefficients, determine if the given #s are roots of the corresponding given equation.<br />6x2 – 5x + 3 = 0(1/6, -1)<br />x2 + x _ 182 = 0(13, -4)<br />x2 – 5x – 14 =0(2, -7)<br />2x – 9x + 8 = 0(3, 4/3)<br />3x2 – 5x – 2 = 0(2, -1/3)<br />Given one roots of the equation, find the other.<br />x2 – 8x – 9 = 0;r= 1<br />2x2 – 3x – 9 = 0;r= 3<br />x2 + x – 2 - 2 = 0; r= 2<br />

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