Must you solve the original equation before transforming its roots?

No. The substitution acts on the equation itself, so the transformed equation is found without ever calculating a single root. That is the real value of the method, and it keeps working when the original roots are irrational or complex and could not be written down neatly anyway.

A LevelFurther MathsEdexcelCore PureFlashcardsFurther Algebra & FunctionsRoots of Polynomials

Roots of Polynomials (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontRoots of Polynomials
How are the relationships between the roots of a polynomial and its coefficients obtained?

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How are the relationships between the roots of a polynomial and its coefficients obtained?
Write the polynomial in both forms, the expanded one and the factorised one $a (x - α) (x - β) \dots$ , then divide through by $a$ and compare coefficients of each power of $x$ .
Expanding the factorised form makes every coefficient a symmetric expression in the roots, which is what produces the relationships.
For the quadratic equation $a x^{2} + b x + c = 0$ with roots $α$ and $β$ , fill in the two missing expressions:
$α + β =______and α β =______$
The completed relationships are:
$α + β = - \frac{b}{a} and α β = \frac{c}{a}$
For a monic quadratic, where $a = 1$ , these are simply $- b$ and $c$ .
You have worked out the sum and the product of the roots of a new quadratic. How do you write down the quadratic itself?
Substitute them into:
$x^{2} - (sum of roots) x + (product of roots) = 0$
The sum enters with a minus sign and the product with a plus, which is exactly how they come out of expanding $(x - α) (x - β)$ .
True or False?
A given pair of roots determines exactly one quadratic equation.
False.
Multiplying an entire equation by any non-zero constant changes every coefficient but leaves the roots untouched, so $2 x^{2} - 5 x - 3 = 0$ and $4 x^{2} - 10 x - 6 = 0$ have the same two roots.
An equation built from a sum and a product of roots is therefore never unique, and you are free to scale it so that the coefficients come out as integers.
For a quadratic with roots $α$ and $β$ , how is $α^{3} + β^{3}$ built from $α + β$ and $α β$ ?
Use the identity:
$α^{3} + β^{3} = {(α + β)}^{3} - 3 α β (α + β)$
Expanding ${(α + β)}^{3}$ gives the two cubes plus $3 α β (α + β)$ , so subtracting that term leaves the cubes on their own.
For a cubic with roots $α$ , $β$ and $γ$ , what does $\sum α β$ mean?
It means the sum of all the products of the roots taken two at a time, which for a cubic is $α β + β γ + α γ$ .
It is a single number built from every such pair, not one individual product.
For the cubic equation $a x^{3} + b x^{2} + c x + d = 0$ , fill in the three missing expressions:
$\sum α =______, \sum α β =______and \sum α β γ =______$
The completed relationships are:
$\sum α = - \frac{b}{a}, \sum α β = \frac{c}{a} and \sum α β γ = - \frac{d}{a}$
Each one uses the next coefficient along the equation, and the signs alternate as you go.
A cubic has roots $α$ , $β$ and $γ$ . How do you find $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$ without first finding the roots?
Put the three fractions over the common denominator $α β γ$ , which gives:
$\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{α β + β γ + γ α}{α β γ} = \frac{\sum α β}{\sum α β γ}$
Both of those are already known from the coefficients, so no root ever has to be calculated.
For a quartic equation, how many different products of the roots taken two at a time are there, and what do they add up to?
There are six of them, since there are six ways to choose two roots from four:
$\sum α β = α β + α γ + α δ + β γ + β δ + γ δ = \frac{c}{a}$
A common slip is to write down only the three neighbouring pairs and lose the other three.
What sign pattern links a polynomial's coefficients to the sums of products of its roots?
The signs alternate, beginning with a minus:
$\sum α = - \frac{b}{a}, \sum α β = \frac{c}{a}, \sum α β γ = - \frac{d}{a}, \dots$
Each sum uses the next coefficient along, and the sum of the products taken $k$ at a time is ${(- 1)}^{k}$ times that coefficient divided by $a$ .
You know $\sum α$ and $\sum α β$ for an equation. How do you find the sum of the squares of its roots?
Use the identity:
$\sum α^{2} = {(\sum α)}^{2} - 2 \sum α β$
Squaring $\sum α$ produces every squared root once and every product of a pair twice, so subtracting $2 \sum α β$ removes the cross terms exactly.
It holds whatever the degree of the equation.
Define a linear transformation of the roots of a polynomial equation.
It is a rule of the form $w = p x + q$ , where $p$ and $q$ are constants, which turns each root $x$ of the original equation into a root $w$ of a new one.
It has the effect of stretching and translating the roots, and $p$ must be non-zero for the new equation to have as many roots as the old.
A linear transformation $w = p x + q$ is to be substituted into the original equation. What has to be done to it first, and why?
Rearrange it to make the original variable the subject, giving $x = \frac{w - q}{p}$ .
The original equation is an equation in $x$ , so it is every $x$ that has to be replaced; substituting $p x + q$ in place of $w$ instead leaves the equation untouched.
The equation $x^{3} - 7 x^{2} + 2 x + 40 = 0$ has roots $5$ , $- 2$ and $4$ . Fill in the roots of the equation produced by the transformation $w = 2 x - 1$ :
$w =______,______and______$
The completed roots are:
$w = 9, - 5 and 7$
The transformation is applied to each root separately, so $2 (5) - 1 = 9$ , $2 (- 2) - 1 = - 5$ and $2 (4) - 1 = 7$ .
You substitute for $x$ in an equation in order to transform its roots. Why does every original root turn into exactly one root of the new equation?
Because the expression substituted is just the transformation rearranged, so the new equation is satisfied precisely when $w$ is the transform of a root of the original.
That makes the correspondence between old roots and new roots one to one, which is why the new equation has the same degree as the old.
You have expanded and simplified the transformed equation in terms of $w$ . What is the final step?
Replace every $w$ with $x$ , since the letter standing for the variable carries no meaning of its own and the answer is conventionally written in $x$ .
It is usual to multiply through by a constant first, so that the coefficients come out as integers rather than fractions.
True or False?
If the roots of a cubic are transformed using $w = 2 x - 1$ , the new equation is also a cubic.
True.
Each $x$ is replaced by an expression that is linear in $w$ , so cubing it still reaches $w^{3}$ and no higher power.
A linear transformation changes the coefficients but never the degree of the equation.
Must you solve the original equation before transforming its roots?
No. The substitution acts on the equation itself, so the transformed equation is found without ever calculating a single root.
That is the real value of the method, and it keeps working when the original roots are irrational or complex and could not be written down neatly anyway.

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