Download presentation

1
**Lesson 9 – Identities and roots of equations**

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 (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 Ax2 – 5Ax – Bx – 5B + C 3x2 + 11x 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 (3x – 4)(x+5) + 38

5
**Substituting in Values**

3x2 + 11x (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 = 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 = = 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 – 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 = 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 α 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 = αβ + 2(α + β) = -8/3 + 2(10/3) = 8 So 22/3 = -b/a & = 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 = & αβ = 15 (α + β)2 = 49 & α2β2 = 225 α2 + 2αβ + β2 = 49 α β2 = 49 α2 + β2 = 19 So the equation is z2 – 19z = 0 Now do Ex 4B pg 104

Similar presentations

OK

Quadratics 3102.3.30 Solve quadratic equations using multiple methods: factoring, graphing, quadratic formula, or square root principle.

Quadratics 3102.3.30 Solve quadratic equations using multiple methods: factoring, graphing, quadratic formula, or square root principle.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on agriculture and its types Ppt on cloud computing services Ppt on kingdom monera classification Ppt on multipurpose river valley projects Ppt on regular expression test Ppt on earth hour video Ppt on 5 electrical appliances Ppt on central limit theorem formula Ppt on email etiquettes presentation folder Ppt on depth first search graph