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

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

2 Wiltshire 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. 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 a 2 – b 2 (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 : x 2 – 5x + 6 = 0, only has two values that work, 2 & 3. What would happen if you tried to solve the identity (x+3) 2 x 2 + 6x + 9

3 Wiltshire Equating Co-efficient Sometimes you will be given an identity with unknown constants on one side, such as 3x x + 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 Wiltshire Equating Coefficient 3x x + 18 (Ax – B)(x+5) + C If you multiply out the right hand side you get 3x x + 18 Ax 2 – 5Ax – Bx – 5B + C 3x x + 18 Ax 2 – (5A – B)x – 5B + C As both expressions are identically equal then we can equate the Coefficients. The terms in front of the x 2 are equal so: A = 3 The terms in front of the x will be equal so:5A – B = – B = 11 B = 4 Finally the constants at the ends of both equations must be equal: -5B + C = C = 18 C = 38 Now 3x x + 18 (3x – 4)(x+5) + 38

5 Wiltshire Substituting in Values 3x x + 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) (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 Wiltshire Substituting in Values 3x x + 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 = = 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 = -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 – = 6A A = 3 Note Identities are not always written using ` `. Eg sin 2 θ + cos 2 θ = 1 Now do Ex 4A pg 100

7 Wiltshire 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)x 2 + 7x + 12 = 0 b)x 2 – 5x + 6 = 0 c)x 2 + x – 20 = 0 d)2x 2 – 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 Wiltshire Properties of the roots of polynomial equations az 2 + bz + c = 0 a(z - α)(z - β) = 0a = 0 This gives the identity az 2 + bz + c = a(z - α)(z - β) Multiplying out az 2 + bz + c = a(z 2 – αz – βz + αβ) = az 2 – aαz – aβz + aαβ = az 2 – az(α + β) + aαβ Equating coefficients b = – a(α + β)c = aαβ -b/a = α + βc/a = αβ

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

10 Wiltshire Properties of the roots of polynomial equations Find a quadratic equation with roots 2 & -5 -b/a = α + βc/a = αβ -b/a = c/a = -5 × 2 -b/a = -3c/a = -10 Taking a = 1 gives b = 3 & c = -10 So z 2 + 3z – 10 = 0 Note: There are infinitely many solutions to this problem. Taking a = 2 would lead to the equation 2z 2 + 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 Wiltshire Properties of the roots of polynomial equations The roots of the equation 3z 2 – 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 z 2 – 10z – 24 = 0

12 Wiltshire 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 3z 2 – 22z + 24 = 0 Properties of the roots of polynomial equations

13 Wiltshire Properties of the roots of polynomial equations The roots of the equation z 2 – 7z + 15 = 0 are α and β. Find the quadratic equation with roots α 2 and β 2 α + β = 7 & αβ = 15 (α + β) 2 = 49& α 2 β 2 = 225 α 2 + 2αβ + β 2 = 49 α β 2 = 49 α 2 + β 2 = 19 So the equation is z 2 – 19z = 0 Now do Ex 4B pg 104


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