Roots of Quadratic Equations (Edexcel IAL Further Maths)

Sums & Products of Roots

Spec point — 2.1

How do I find the sum and product of the roots of a quadratic?

• The quadratic equation $ax^{2}+bx+c=0$ can be written as $x^{2}+\frac{b}{a}x+\frac{c}{a}=0$

  • by dividing by $a$ (assuming $a\neq0$)

• If $α$ and $β$ are roots of the quadratic equation

  • then the equation is also $( x - \\alpha ) ( x - \\beta ) = 0$ in factorised form

    • This works for real and complex roots

• Both left-hand sides of the equations are identical

  • $( x - \\alpha ) ( x - \\beta ) \equiv x^{2} + \frac{b}{a} x + \frac{c}{a}$

  • Expand and collect the left-hand side

  • $x^{2} - ( \\alpha + \\beta ) x + \\alpha \\beta \equiv x^{2} + \frac{b}{a} x + \frac{c}{a}$

  • Equating coefficients gives

    • the sum of the roots is $α+β=-\frac{b}{a}$
    • the product of the roots is $αβ=\frac{c}{a}$

• Be careful: the sum of the roots has a negative sign

  • The product does not

Exam Hint These two relationships are not given in the Formulae Booklet, so must be learnt!

Worked Example The quadratic equation $5x^{2}-30x-3=0$ has roots $α$ and $β$.

Without solving the equation, find the value of:

(a) $α+β$

Either use $\\alpha + \\beta = - \frac{b}{a} = - \frac{( - 30 )}{5}$ Or first divide both sides of the equation by 5

$x^{2}-6x-\frac{3}{5}=0$

Then use $x ^{2} - ( \text{sum} of \text{roots} ) x + \text{product} of \text{roots} = 0$ The sum of the roots is the negative of the middle number

$\\alpha + \\beta = - ( - 6 )$

$α+β=6$

(b) $αβ$

Either use $\\alpha \\beta = \frac{c}{a} = \frac{( - 3 )}{5}$ Or first divide both sides of the equation by 5

$x^{2}-6x-\frac{3}{5}=0$

Then use $x ^{2} - ( \text{sum} of \text{roots} ) x + \text{product} of \text{roots} = 0$ The product of the roots is the constant term on the end No change of sign is needed

$αβ=-\frac{3}{5}$

Expressions with Roots

Spec point — 2.2

How do I simplify expressions involving roots of a quadratic?

• You can use algebra to write expressions in terms of $( \\alpha + \\beta )$ and $αβ$

• Then use the fact that the equation $ax^{2}+bx+c=0$ has roots $α$ and $β$ where

  • $α+β=-\frac{b}{a}$

  • $αβ=\frac{c}{a}$

    • Substitute these into your expression to find its value

Which identities do I need to know?

• You should know identities involving powers of products of roots

  • $α^{2}β^{2}\equiv(αβ)^{2}$
  • $α^{3}β^{3}\equiv(αβ)^{3}$
  • and so on

• You should know the identity for the sum of the squares of the roots

  • $α^{2}+β^{2}\equiv(α+β)^{2}-2αβ$

    • which comes from expanding $(α+β)^{2}\equivα^{2}+β^{2}+2αβ$

• You should know the identity for the sum of the cubes of the roots

  • $α^{3}+β^{3}\equiv(α+β)^{3}-3αβ(α+β)$

    • which comes from the binomial expansion of $( \\alpha + \\beta )^{3}$
    • $( \\alpha + \\beta )^{3} \equiv \\alpha ^{3} + 3 \\alpha ^{2} \\beta + 3 \\alpha \\beta ^{2} + \\beta ^{3} \equiv \\alpha ^{3} + \\beta ^{3} + 3 \\alpha \\beta ( \\alpha + \\beta )$
    • then rearranging

• You should know how to use algebraic fractions (but do not need to learn these identities)

  • $\frac{1}{α}\times\frac{1}{β}\equiv\frac{1}{αβ}$
  • $\frac{1}{α}+\frac{1}{β}\equiv\frac{α+β}{αβ}$
  • $\frac{α}{β}+\frac{β}{α}\equiv\frac{α^{2}+β^{2}}{αβ}$
  • and so on

Exam Hint None of the identities are given in the Formulae Booklet, so you either need to learn them, or learn how to find them.

Worked Example The quadratic equation $2x^{2}+10x+9=0$ has roots $α$ and $β$.

Without solving the equation, find the exact value of

(a) $\frac{α}{β}+\frac{β}{α}$

Use $α+β=-\frac{b}{a}$ and $αβ=\frac{c}{a}$ to find the sum and product of the roots

$α+β=-\frac{10}{2}=-5$* * and $αβ=\frac{9}{2}$

Add the algebraic fractions $\frac{α}{β}+\frac{β}{α}$ Use a lowest common denominator of $αβ$

| α / β+β / α |=(α ^2)/(α β)+(β ^2)/(α β) | | _ _ _ | _=(α ^2+β ^2)/(α β) | |

Now use the identity for the sum of the squares of the roots $α^{2}+β^{2}\equiv(α+β)^{2}-2αβ$

$= \frac{( \\alpha + \\beta )^{2} - 2 \\alpha \\beta }{\\alpha \\beta }$

Substitute in $α+β=-5$ and $αβ=\frac{9}{2}$

| α / β+β / α |=((- 5) ^2 - 2 (9 / 2))/((9 / 2)) | | _=(25 - 9)/(9 / 2) | | _=16 / 9 / 2 | |

$\begin{array}{ccc}\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } & = & \frac{32}{9}\end{array}$

(b) $α^{3}+β^{3}$

Either use $α^{3}+β^{3}\equiv(α+β)^{3}-3αβ(α+β)$ if learnt Or start by expanding the binomial $(α+β)^{3}$

$(α+β)^{3}=α^{3}+3α^{2}β+3αβ^{2}+β^{3}$

Factorise fully the middle two terms

$( \\alpha + \\beta )^{3} = \\alpha ^{3} + 3 \\alpha \\beta ( \\alpha + \\beta ) + \\beta ^{3}$

Make $α^{3}+β^{3}$ the subject

$\\alpha ^{3} + \\beta ^{3} = ( \\alpha + \\beta )^{3} - 3 \\alpha \\beta ( \\alpha + \\beta )$

Substitute in $α+β=-5$ and $αβ=\frac{9}{2}$

$\begin{gathered} \\alpha ^{3} + \\beta ^{3} & = ( - 5 )^{3} - 3 ( \frac{9}{2} ) ( - 5 ) \\ = - 125 + \frac{135}{2} \end{gathered}$

$α^{3}+β^{3}=-\frac{115}{2}$

-57.5 is also accepted