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Further Pure 1

Further Pure 1. Lesson 9 – Identities and roots of equations. Identities. x 3 – y 3 (x-y)(x 2 + xy + y 2 ) is an example of an identity. The 3 lined equals sign means identically equal to.

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Further Pure 1

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  1. Further Pure 1 Lesson 9 – Identities and roots of equations

  2. Identities • x3 – y3 (x-y)(x2 + xy + y2) is an example of an identity. • The 3 lined equals sign means identically equal to. • This means that both sides of the equation will always be equal whatever the values of x and y are. • Here are some more examples of identities 2(x+3) 2x + 6 a2 – b2 (a+b)(a-b) • In an identity all possible values of the variable(s) will satisfy the identity. • With an equation only certain values satisfy the equation. • Example : x2 – 5x + 6 = 0, only has two values that work, 2 & 3. • What would happen if you tried to solve the identity (x+3)2 x2 + 6x + 9

  3. Equating Co-efficient • Sometimes you will be given an identity with unknown constants on one side, such as 3x2 + 11x + 18 (Ax – B)(x+5) + C • There are two methods to finding out the values of these constants. • Method 1 – Equating Coefficients • Method 2 – Substituting in values

  4. Equating Coefficient • 3x2 + 11x + 18 (Ax – B)(x+5) + C • If you multiply out the right hand side you get 3x2 + 11x + 18 Ax2 – 5Ax – Bx – 5B + C 3x2 + 11x + 18 Ax2 – (5A – B)x – 5B + C • As both expressions are identically equal then we can equate the Coefficients. • The terms in front of the x2 are equal so: A = 3 • The terms in front of the x will be equal so: 5A – B = 11 15 – B = 11 B = 4 • Finally the constants at the ends of both equations must be equal: -5B + C = 18 -20 + C = 18 C = 38 • Now 3x2 + 11x + 18 (3x – 4)(x+5) + 38

  5. Substituting in Values • 3x2 + 11x + 18 (Ax – B)(x+5) + C • Since the expressions are equal we can plug in any values we like for x to form equations in A,B and C that we can solve. • Example if x = 1, then 3(1)2 + 11(1) + 18 = (A(1) – B)(1 + 5) + C 32 = 6A - 6B + C • The problem is though that you now have one equation with 3 unknowns. • When you pick your value of x to plug in try to pick values that will cancel out some of the unknowns.

  6. Substituting in Values • 3x2 + 11x + 18 = (Ax – B)(x+5) + C • If you pick x = -5 then all of the expression (Ax – B)(x+5) is equal to zero. • So C = 75 - 55 + 18 = 38 • Now if x = 0 then the A term will go and we are left with B and C, however we already know what C is. • So 3(0) + 11(0) + 18 = (A(0) – B)(0 + 5) + C 18 = -5B + 38 -20 = -5B B = 4 • Finally we could use the example of x = 1 on the previous slide because we now know B & C. • If x = 1, 32 = 6A – 6B + C 32 = 6A – 24 + 38 18 = 6A A = 3 • Note Identities are not always written using ` `. Eg sin2θ + cos2θ = 1 • Now do Ex 4A pg 100

  7. Properties of the roots of polynomial equations • In this chapter the variable x is replaced with a z to emphasize that the results could be complex or real. • 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)

  8. Properties of the roots of polynomial equations • az2 + bz + c = 0 a(z - α)(z - β) = 0 a = 0 • This gives the identity az2 + bz + c = a(z - α)(z - β) • Multiplying out az2 + bz + c = a(z2 – αz – βz + αβ) = az2 – aαz – aβz + aαβ = az2 – az(α + β) + aαβ • Equating coefficients b = – a(α + β) c = aαβ -b/a = α + β c/a = αβ

  9. Task • Use the quadratic formula to prove the results from the previous slide. -b/a = α + β c/a = αβ

  10. Properties of the roots of polynomial equations • 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 z2 + 3z – 10 = 0 • Note: There are infinitely many solutions to this problem. • Taking a = 2 would lead to the equation 2z2 + 6z – 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.

  11. Properties of the roots of polynomial equations • The roots of the equation 3z2 – 10z – 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 z2 – 10z – 24 = 0

  12. Properties of the roots of polynomial equations 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 3z2 – 22z + 24 = 0

  13. Properties of the roots of polynomial equations • The roots of the equation z2 – 7z + 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 z2 – 19z + 225 = 0 • Now do Ex 4B pg 104

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