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Roots of Quadratic Equations

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Roots of Quadratic Equations

  1. 1. Investigating the relationships between the roots and the coefficients of quadratic equations. T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  2. 2.  Solve each of the following quadratic equations a) x2 + 7x + 12 = 0 b) x2 – 5x + 6 = 0 c) x2 + x – 20 = 0 d) 2x2 – 5x – 3 = 0  Write down the sum of the roots and the product of the roots.  Roots of polynomial equations are usually denoted by Greek letters.  For a quadratic equation we use alpha (α) & beta (β)
  3. 3.  ax2 + bx + c = 0 a(x - α)(x - β) = 0 a = 0  This gives the identity ax2 + bx + c = a(x - α)(x - β)  Multiplying out ax2 + bx + c = a(x2 – αx – βx + αβ) = ax2 – aαx – aβx + aαβ = ax2 – ax(α + β) + aαβ  Equating coefficients b = – a(α + β) c = aαβ -b/a = α + β c/a = αβ
  4. 4.  Use the quadratic formula to prove the results from the previous slide. -b/a = α + β c/a = αβ a acbb 2 42 a acbb 2 42 a b a b a acbb a acbb 2 2 2 4 2 4 22 a c a ac a acbb a acbb a acbb 22 2222 4 4 4 )4( 2 4 2 4
  5. 5.  Find a quadratic equation with roots 2 & -5 -b/a = α + β c/a = αβ -b/a = 2 + -5 c/a = -5 2 -b/a = -3 c/a = -10  Taking a = 1 gives b = 3 & c = -10  So x2 + 3x – 10 = 0  Note: There are infinitely many solutions to this problem.  Taking a = 2 would lead to the equation 2x2 + 6x – 20 = 0  Taking a = 1 gives us the easiest solution.  If b and c are fractions you might like to pick an appropriate value for a.
  6. 6.  The roots of the equation 3x2 – 10x – 8 = 0 are α & β 1 – Find the values of α + β and αβ. α + β = -b/a = 10/3 αβ = c/a = -8/3 2 – Find the quadratic equation with roots 3α and 3β.  The sum of the new roots is 3α + 3β = 3(α + β) = 3 10/3 = 10  The product of the new roots is 9αβ = -24  From this we get that 10 = -b/a & -24 = c/a  Taking a = 1 gives b = -10 & c = -24  So the equation is x2 – 10x – 24 = 0
  7. 7. 3 – Find the quadratic equation with roots α + 2 and β + 2  The sum of the new roots is α + β + 4 = 10/3 + 4 = 22/3  The product of the new roots is (α + 2)(β + 2) = αβ + 2α + 2β + 4 = αβ + 2(α + β) + 4 = -8/3 + 2(10/3) + 4 = 8  So 22/3 = -b/a & 8 = c/a  To get rid of the fraction let a = 3, so b = -22 & c = 24  The equation is 3x2 – 22x + 24 = 0
  8. 8.  The roots of the equation x2 – 7x + 15 = 0 are α and β.  Find the quadratic equation with roots α2 and β2 α + β = 7 & αβ = 15 (α + β)2 = 49 & α2β2 = 225 α2 + 2αβ + β2 = 49 α2 + 30 + β2 = 49 α2 + β2 = 19  So the equation is x2 – 19x + 225 = 0
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