In General Maths you met the sum and product of the roots (see quadratic equations↺ ). If α \alpha α β \beta β a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0
α + β = − b a α β = c a \alpha + \beta = -\frac ba \qquad \alpha\beta = \frac ca α + β = − a b α β = a c Further Maths questions use these two facts to find other things about the roots, often when the roots themselves are awkward surds. This lesson shows where the facts come from, then uses them.
Where the two facts come from
If α \alpha α β \beta β ( x − α ) (x - \alpha) ( x − α ) ( x − β ) (x - \beta) ( x − β )
a x 2 + b x + c = a ( x − α ) ( x − β ) ax^2 + bx + c = a(x - \alpha)(x - \beta) a x 2 + b x + c = a ( x − α ) ( x − β )
Multiply out the brackets: ( x − α ) ( x − β ) = x 2 − α x − β x + α β {(x - \alpha)(x - \beta) = x^2 - \alpha x - \beta x + \alpha\beta} ( x − α ) ( x − β ) = x 2 − α x − β x + α β
Collect the x x x x 2 − ( α + β ) x + α β {x^2 - (\alpha + \beta)x + \alpha\beta} x 2 − ( α + β ) x + α β
Multiply by a a a a x 2 − a ( α + β ) x + a α β {ax^2 - a(\alpha + \beta)x + a\alpha\beta} a x 2 − a ( α + β ) x + a α β
Match the x x x b x bx b x b = − a ( α + β ) {b = -a(\alpha + \beta)} b = − a ( α + β ) α + β = − b a {\alpha + \beta = -\frac ba} α + β = − a b
Match the constants with c c c c = a α β {c = a\alpha\beta} c = a α β α β = c a {\alpha\beta = \frac ca} α β = a c
ax² + bx + c = 0, roots α and β
α + β = −b⁄a αβ = c⁄a
The equation is x² − (sum )x + (product ) = 0
Sum and product of the roots Sum −b/a, product c/a More: the sum and product
Expressions in α and β
Many expressions stay the same when you swap α \alpha α β \beta β α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2 α + β \alpha + \beta α + β α β \alpha\beta α β
The key one comes from squaring the sum. The square of side α + β \alpha + \beta α + β α 2 \alpha^2 α 2 β 2 \beta^2 β 2 α β \alpha\beta α β
α² αβ αβ β² α β α β (α + β)² = α² + 2αβ + β² Take away the two rectangles: α² + β² = (α + β)² − 2αβ
The others you will need:
1 α + 1 β = α + β α β ( α − β ) 2 = ( α + β ) 2 − 4 α β α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) \begin{aligned}
\frac1\alpha + \frac1\beta &= \frac{\alpha + \beta}{\alpha\beta} \\
(\alpha - \beta)^2 &= (\alpha + \beta)^2 - 4\alpha\beta \\
\alpha^3 + \beta^3 &= (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)
\end{aligned} α 1 + β 1 ( α − β ) 2 α 3 + β 3 = α β α + β = ( α + β ) 2 − 4 α β = ( α + β ) 3 − 3 α β ( α + β ) Each comes from multiplying out. For example, for the cube:
Multiply out: ( α + β ) 3 = α 3 + 3 α 2 β + 3 α β 2 + β 3 {(\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3} ( α + β ) 3 = α 3 + 3 α 2 β + 3 α β 2 + β 3
Take out 3 α β 3\alpha\beta 3 α β = α 3 + β 3 + 3 α β ( α + β ) {= \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)} = α 3 + β 3 + 3 α β ( α + β )
Take 3 α β ( α + β ) 3\alpha\beta(\alpha + \beta) 3 α β ( α + β ) α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) {\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)} α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β )
Pick an equation and an expression, and compare the answer with the actual roots:
Sum and product at work Pick an equation and an expression
α² + β² = (α + β)² − 2αβ = 5² − 2 × 3 = 25 − 6 = 19 −1 1 2 3 4 5 6 −4 −2 2 4 6 x y α β 5 α + β = −b/a 3 αβ = c/a 19 α² + β²
x² − 5x + 3 = 0 2x² + 3x − 4 = 0 3x² − 6x − 2 = 0 α² + β² 1/α + 1/β (α − β)² α³ + β³
The roots themselves are awkward surds, about 0.697 and 4.303. Working with them directly gives α² + β² ≈ 19: the same answer as the sum and product give exactly, without ever solving the equation. For example, for 2 x 2 − 4 x − 3 = 0 2x^2 - 4x - 3 = 0 2 x 2 − 4 x − 3 = 0
The sum: α + β = − − 4 2 = 2 {\alpha + \beta = -\frac{-4}{2} = 2} α + β = − 2 − 4 = 2
The product: α β = − 3 2 {\alpha\beta = \frac{-3}{2}} α β = 2 − 3
So α 2 + β 2 = 2 2 − 2 ( − 3 2 ) {\alpha^2 + \beta^2 = 2^2 - 2\left(-\frac32\right)} α 2 + β 2 = 2 2 − 2 ( − 2 3 )
= 4 + 3 = 7 {= 4 + 3 = 7} = 4 + 3 = 7
Worked example · WAEC 2022
WAEC 2022 · Paper 2 · Q2
If α \alpha α β \beta β 2 x 2 − x − 2 = 0 2x^2 - x - 2 = 0 2 x 2 − x − 2 = 0 α 3 + β 3 \alpha^3 + \beta^3 α 3 + β 3
Value of α 3 + β 3 \alpha^3 + \beta^3 α 3 + β 3
The sum and the product
α + β = − − 1 2 = 1 2 {\alpha + \beta = -\frac{-1}{2} = \frac12} α + β = − 2 − 1 = 2 1 α β = − 2 2 = − 1 {\alpha\beta = \frac{-2}{2} = -1} α β = 2 − 2 = − 1 Think first. a = 2, b = −1, c = −2. What are α + β and αβ?
Write the cube in terms of them
α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) Substitute
( 1 2 ) 3 = 1 8 {\left(\frac12\right)^3 = \frac18} ( 2 1 ) 3 = 8 1 3 α β ( α + β ) = 3 ( − 1 ) ( 1 2 ) = − 3 2 {3\alpha\beta(\alpha + \beta) = 3(-1)\left(\frac12\right) = -\frac32} 3 α β ( α + β ) = 3 ( − 1 ) ( 2 1 ) = − 2 3 So α 3 + β 3 = 1 8 − ( − 3 2 ) {\alpha^3 + \beta^3 = \frac18 - \left(-\frac32\right)} α 3 + β 3 = 8 1 − ( − 2 3 )
= 1 8 + 12 8 = 13 8 {= \frac18 + \frac{12}{8} = \frac{13}{8}} = 8 1 + 8 12 = 8 13 Think first. Put in ½ and −1.
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For α − β \alpha - \beta α − β ± \pm ± α \alpha α
More: expressions in α and β
An equation with new roots
Any quadratic with roots p p p q q q
x 2 − ( p + q ) x + p q = 0 x^2 - (p + q)x + pq = 0 x 2 − ( p + q ) x + pq = 0 so all you need for the new equation is the sum and the product of the new roots. Work each one out in terms of α + β \alpha + \beta α + β α β \alpha\beta α β
Worked example · WAEC 2011
WAEC 2011 · Paper 2 · Q2
If α \alpha α β \beta β 2 x 2 − 7 x + 4 = 0 2x^2 - 7x + 4 = 0 2 x 2 − 7 x + 4 = 0 α β \dfrac\alpha\beta β α β α \dfrac\beta\alpha α β
The old sum and product
α + β = 7 2 {\alpha + \beta = \frac72} α + β = 2 7 α β = 4 2 = 2 {\alpha\beta = \frac42 = 2} α β = 2 4 = 2 Think first. For 2x² − 7x + 4 = 0, what are α + β and αβ?
The sum of the new roots
Add the fractions: α β + β α = α 2 + β 2 α β {\frac\alpha\beta + \frac\beta\alpha = \frac{\alpha^2 + \beta^2}{\alpha\beta}} β α + α β = α β α 2 + β 2
The top: α 2 + β 2 = ( 7 2 ) 2 − 2 ( 2 ) {\alpha^2 + \beta^2 = \left(\frac72\right)^2 - 2(2)} α 2 + β 2 = ( 2 7 ) 2 − 2 ( 2 )
= 49 4 − 16 4 = 33 4 {= \frac{49}{4} - \frac{16}{4} = \frac{33}{4}} = 4 49 − 4 16 = 4 33 Divide by α β = 2 \alpha\beta = 2 α β = 2 33 8 {\frac{33}{8}} 8 33
Think first. Add α/β and β/α as fractions.
The product of the new roots
α β × β α = 1 {\frac\alpha\beta \times \frac\beta\alpha = 1} β α × α β = 1 Think first. What is α/β × β/α?
The equation
x 2 − 33 8 x + 1 = 0 {x^2 - \frac{33}{8}x + 1 = 0} x 2 − 8 33 x + 1 = 0 Multiply through by 8: 8 x 2 − 33 x + 8 = 0 {8x^2 - 33x + 8 = 0} 8 x 2 − 33 x + 8 = 0
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More: equations with new roots
Your turn
(b) If α \alpha α β \beta β 3 x 2 − 5 x + 1 = 0 3x^2 - 5x + 1 = 0 3 x 2 − 5 x + 1 = 0 ( α − 1 β ) \left(\alpha - \frac1\beta\right) ( α − β 1 ) ( β − 1 α ) \left(\beta - \frac1\alpha\right) ( β − α 1 )
Show the answer 3 x 2 + 10 x + 4 = 0 3x^2 + 10x + 4 = 0 3 x 2 + 10 x + 4 = 0
Worked solution (try it first) (b) For
3 x 2 − 5 x + 1 = 0 3x^2 - 5x + 1 = 0 3 x 2 − 5 x + 1 = 0 :
α + β = 5 3 \alpha + \beta = \frac53 α + β = 3 5 and
α β = 1 3 \alpha\beta = \frac13 α β = 3 1 .
New sum:
( α − 1 β ) + ( β − 1 α ) = ( α + β ) − α + β α β \left(\alpha - \frac1\beta\right) + \left(\beta - \frac1\alpha\right) = (\alpha + \beta) - \dfrac{\alpha + \beta}{\alpha\beta} ( α − β 1 ) + ( β − α 1 ) = ( α + β ) − α β α + β .
Substitute:
5 3 − 5 3 1 3 = 5 3 − 5 \frac53 - \dfrac{\frac53}{\frac13} = \frac53 - 5 3 5 − 3 1 3 5 = 3 5 − 5 = − 10 3 = -\frac{10}{3} = − 3 10 .
New product:
( α − 1 β ) ( β − 1 α ) = α β − 1 − 1 + 1 α β \left(\alpha - \frac1\beta\right)\left(\beta - \frac1\alpha\right) = \alpha\beta - 1 - 1 + \dfrac{1}{\alpha\beta} ( α − β 1 ) ( β − α 1 ) = α β − 1 − 1 + α β 1 .
Substitute:
1 3 − 2 + 3 = 4 3 \frac13 - 2 + 3 = \frac43 3 1 − 2 + 3 = 3 4 .
The equation:
x 2 + 10 3 x + 4 3 = 0 x^2 + \frac{10}{3}x + \frac43 = 0 x 2 + 3 10 x + 3 4 = 0 .
Multiply through by 3:
3 x 2 + 10 x + 4 = 0 3x^2 + 10x + 4 = 0 3 x 2 + 10 x + 4 = 0 .
Watch out
In (a), 3 x y 3xy 3 x y 3 y + 3 x d y d x 3y + 3x\dfrac{dy}{dx} 3 y + 3 x d x d y 3 d y d x 3\dfrac{dy}{dx} 3 d x d y In (b), α β = c a = + 1 3 \alpha\beta = \frac ca = +\frac13 α β = a c = + 3 1 α × β \alpha \times \beta α × β α × ( − 1 α ) \alpha \times \left(-\frac1\alpha\right) α × ( − α 1 ) − 1 β × β -\frac1\beta \times \beta − β 1 × β 1 α β \frac{1}{\alpha\beta} α β 1 Report a problem with this question