Forming Quadratics with New Roots (Edexcel International A Level Further Maths)

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Author

Mark Curtis

Expertise

Maths

Forming Quadratics with New Roots

How do I form quadratics with new roots?

The quadratic equation $a x^{2} + b x + c = 0$ has roots $α$ and $β$ where
- $α + β = - \frac{b}{a}$
- $α β = \frac{c}{a}$
Any new quadratic equations with new roots will have the form
- $x^{2} - (sum of roots) x + (product of roots) = 0$
For example, to find the quadratic equation with roots $α^{2}$ and $β^{2}$
- The new equation is $x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2} = 0$
- You can then work out these coefficients using the identities from before
  - $α^{2} + β^{2} \equiv {(α + β)}^{2} - 2 α β$ and $α^{2} β^{2} \equiv {(α β)}^{2}$
  - Substitute in $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$

Which identities do I need to know?

You should know identities involving powers of products of roots
- $α^{2} β^{2} \equiv (α β)^{2}$
- $α^{3} β^{3} \equiv (α β)^{3}$
- and so on
You should know the identity for the sum of the squares of the roots
- $α^{2} + β^{2} \equiv (α + β)^{2} - 2 α β$
  - which comes from expanding $(α + β)^{2} \equiv α^{2} + β^{2} + 2 α β$
You should know the identity for the sum of the cubes of the roots
- $α^{3} + β^{3} \equiv (α + β)^{3} - 3 α β (α + β)$
  - which comes from the binomial expansion of ${(α + β)}^{3}$
  - $(α + β)^{3} \equiv α^{3} + 3 α^{2} β + 3 α β^{2} + β^{3} \equiv α^{3} + β^{3} + 3 α β (α + β)$
  - then rearranging

You should know how to use algebraic fractions (but do not need to learn these identities)
- $\frac{1}{α} \times \frac{1}{β} \equiv \frac{1}{α β}$
- $\frac{1}{α} + \frac{1}{β} \equiv \frac{α + β}{α β}$
- $\frac{α}{β} + \frac{β}{α} \equiv \frac{α^{2} + β^{2}}{α β}$
- and so on

Exam Tip

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 $3 x^{2} + 12 x - 1 = 0$ 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 $A x^{2} + B x + C = 0$ where $A$ , $B$ and $C$ are integers to be determined.

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

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

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

$x^{2} - (\frac{β}{α} - α + \frac{α}{β} - β) x + (\frac{β}{α} - α) (\frac{α}{β} - β) = 0$

Simplify the coefficient of $x$
Add the algebraic fractions

$\begin{array}{rcl} \frac{β}{α} - α + \frac{α}{β} - β & = & \frac{β}{α} + \frac{α}{β} - (α + β) \\ = & \frac{β^{2} + α^{2}}{α β} - (α + β) \end{array}$

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

$\begin{array}{rcl} = & \frac{(α + β)^{2} - 2 α β}{α β} - (α + β) \end{array}$

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

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

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

$\begin{array}{rcl} (\frac{β}{α} - α) (\frac{α}{β} - β) & = & \frac{β α}{α β} - \frac{α^{2}}{β} - \frac{β^{2}}{α} + α β \\ = & 1 - (\frac{α^{2}}{β} + \frac{β^{2}}{α}) + α β \\ = & 1 - \frac{(α^{3} + β^{3})}{α β} + α β \end{array}$

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

$\begin{array}{rcl} = & 1 - \frac{[(α + β)^{3} - 3 α β (α + β)]}{α β} + α β \end{array}$

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

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

Write out the new quadratic equation
Simplify the coefficients

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

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

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

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

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

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