Presentation on theme: "Lesson 9 – Identities and roots of equations"— Presentation transcript:
1 Lesson 9 – Identities and roots of equations Further Pure 1Lesson 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 identities2(x+3) 2x + 6a2 – 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 (Ax – B)(x+5) + CThere are two methods to finding out the values of these constants.Method 1 – Equating CoefficientsMethod 2 – Substituting in values
4 Equating Coefficient 3x2 + 11x + 18 (Ax – B)(x+5) + C If you multiply out the right hand side you get3x2 + 11x Ax2 – 5Ax – Bx – 5B + C3x2 + 11x Ax2 – (5A – B)x – 5B + CAs both expressions are identically equal then we can equate the Coefficients.The terms in front of the x2 are equal so: A = 3The terms in front of the x will be equal so: 5A – B = 1115 – B = 11B = 4Finally the constants at the ends of both equations must be equal:-5B + C = 18-20 + C = 18C = 38Now 3x2 + 11x (3x – 4)(x+5) + 38
5 Substituting in Values 3x2 + 11x (Ax – B)(x+5) + CSince 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 = 6A - 6B + CThe 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) + CIf you pick x = -5 then all of the expression (Ax – B)(x+5) is equal to zero.So C = = 38Now 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) + C18 = -5B + 38-20 = -5BB = 4Finally we could use the example of x = 1 on the previous slide because we now know B & C.If x = 1, 32 = 6A – 6B + C32 = 6A –18 = 6AA = 3Note Identities are not always written using ` `. Eg sin2θ + cos2θ = 1Now 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 equationsa) x2 + 7x + 12 = 0b) x2 – 5x + 6 = 0c) x2 + x – 20 = 0d) 2x2 – 5x – 3 = 0Write 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 = 0a(z - α)(z - β) = 0 a = 0This gives the identityaz2 + bz + c = a(z - α)(z - β)Multiplying outaz2 + bz + c = a(z2 – αz – βz + αβ)= az2 – aαz – aβz + aαβ= az2 – az(α + β) + aαβEquating coefficientsb = – a(α + β) c = aαβ-b/a = α + β c/a = αβ
9 TaskUse 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 = c/a = -5 × 2-b/a = -3 c/a = -10Taking a = 1 gives b = 3 & c = -10So z2 + 3z – 10 = 0Note: There are infinitely many solutions to this problem.Taking a = 2 would lead to the equation 2z2 + 6z – 20 = 0Taking 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 & β1 – Find the values of α + β and αβ.α + β = -b/a = 10/3αβ = c/a = -8/32 – Find the quadratic equation with roots α and 3β.The sum of the new roots is 3α + 3β = 3(α + β) = 3 × 10/3 = 10The product of the new roots is 9αβ = -24From this we get that 10 = -b/a & -24 = c/aTaking a = 1 gives b = -10 & c = -24So 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 + 2The sum of the new roots is α + β + 4 = 10/3 + 4 = 22/3The product of the new roots is (α + 2)(β + 2) = αβ + 2α + 2β = αβ + 2(α + β) = -8/3 + 2(10/3) = 8So 22/3 = -b/a & = c/aTo get rid of the fraction let a = 3, so b = -22 & c = 24The 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 β2α + β = & αβ = 15(α + β)2 = 49 & α2β2 = 225α2 + 2αβ + β2 = 49α β2 = 49α2 + β2 = 19So the equation is z2 – 19z = 0Now do Ex 4B pg 104
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