Forming Quadratics with New Roots (Edexcel IAL Further Maths)

Spec point — 2.3

Forming Quadratics with New Roots

How do I form quadratics with new roots?

• The quadratic equation $ax^{2}+bx+c=0$ has roots $α$ and $β$ where

  • $α+β=-\frac{b}{a}$
  • $αβ=\frac{c}{a}$

• Any new quadratic equations with new roots will have the form

  • $x ^{2} - ( \text{sum} of \text{roots} ) x + ( \text{product} of \text{roots} ) = 0$

• For example, to find the quadratic equation with roots $α^{2}$ and $β^{2}$

  • The new equation is $x^{2} - ( \\alpha ^{2} + \\beta ^{2} ) x + \\alpha ^{2} \\beta ^{2} = 0$

  • You can then work out these coefficients using the identities from before

    • $\\alpha ^{2} + \\beta ^{2} \equiv ( \\alpha + \\beta )^{2} - 2 \\alpha \\beta$ and $\\alpha ^{2} \\beta ^{2} \equiv ( \\alpha \\beta )^{2}$
    • Substitute in $α+β=-\frac{b}{a}$ and $αβ=\frac{c}{a}$

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

• Don't forget to add $=0$ to your quadratic equations!

• Read the question carefully to see how to give your final answer

  • For example, if asked for integer coefficients

Worked Example The quadratic equation $3x^{2}+12x-1=0$ has roots $α$ and $β$.

Without solving the equation, find a quadratic equation with roots

$\frac{β}{α}-α$ and $\frac{α}{β}-β$

giving your answer in the form $Ax^{2}+Bx+C=0$ where $A$, $B$ and $C$ are integers to be determined.

Start with the equation $3x^{2}+12x-1=0$ Use $α+β=-\frac{b}{a}$ and $αβ=\frac{c}{a}$ to find the sum and product of the roots

$α+β=-\frac{12}{3}=-4$ and $αβ=-\frac{1}{3}$

Form the new equation using $x ^{2} - ( \text{sum} of \text{roots} ) x + ( \text{product} of \text{roots} ) = 0$

$x ^{2} - ( \frac{ \beta }{ \alpha } - \alpha + \frac{ \alpha }{ \beta } - \beta ) x + ( \frac{ \beta }{ \alpha } - \alpha ) ( \frac{ \alpha }{ \beta } - \beta ) = 0$

Simplify the coefficient of $x$ Add the algebraic fractions

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

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

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

Substitute in $α+β=-4$ and $αβ=-\frac{1}{3}$

$\begin{gathered} = \frac{( - 4 )^{2} - 2 ( - 1 \frac{1}{3} )}{( - 1 \frac{1}{3} )} - ( - 4 ) \\ = - 46 \end{gathered}$

Now simplify the constant term Expand the brackets and add the algebraic fractions

| (β / α - α)(α / β - β) |=(β α)/(α β) - α ^2 / β - β ^2 / α+α β | | _=1 - (α ^2 / β+β ^2 / α)+α β | | _=1 - ((α ^3+β ^3))/(α β)+α β | |

Use (or derive) the identity for the sum the cubes of the roots $α^{3}+β^{3}\equiv(α+β)^{3}-3αβ(α+β)$

$\begin{gathered} = 1 - \frac{[ ( \\alpha + \\beta )^{3} - 3 \\alpha \\beta ( \\alpha + \\beta ) ]}{\\alpha \\beta } + \\alpha \\beta \end{gathered}$

Substitute in $α+β=-4$ and $αβ=-\frac{1}{3}$

$\begin{gathered} = 1 - \frac{[ ( - 4 )^{3} - 3 ( - 1 \frac{1}{3} ) ( - 4 ) ]}{( - 1 \frac{1}{3} )} + ( - 1 \frac{1}{3} ) \\ = 1 - 204 - 1 \frac{1}{3} \\ = - \frac{610}{3} \end{gathered}$

Write out the new quadratic equation Simplify the coefficients

$\begin{gathered} x^{2} - ( - 46 ) x + ( - \frac{610}{3} ) & = 0 \\ x^{2} + 46 x - \frac{610}{3} & = 0 \end{gathered}$

The question asks for the equation to have integer coefficients Multiply both sides by 3

$\begin{gathered} 3 x^{2} + 138 x - 610 & = 0 \end{gathered}$

Don't forget to write$=0$to get full marks

Any multiple of the answer (e.g.$\begin{gathered} 6 x^{2} + 276 x - 1220 & = 0 \end{gathered}$) would also be accepted