Quadratic FunctionsQuadratic Functions
Sum and Product of RootsSum and Product of Roots
PurposePurpose
Find sum/product of roots Find sum/product of roots withoutwithout knowing knowing the actual value of x.the actual value of x.
Use the sum/product of roots to solve for Use the sum/product of roots to solve for other resultsother results
Derivation of FormulaDerivation of Formula
Given Given αα and and ββ are roots of a quadratic are roots of a quadratic equation, we can infer thatequation, we can infer that
xx11 = = αα and x and x22 = = ββ , where , where αα and and ββ are just are just
some numbers.some numbers.
For example, xFor example, x22 – 5x + 6 = 0 has roots – 5x + 6 = 0 has roots αα and and ββ means that means that αα = 2 and = 2 and ββ = 3 = 3
However…However…
Sometimes, we do not need the values of x Sometimes, we do not need the values of x to help us solve the problem. to help us solve the problem.
Knowing the relationship between the Knowing the relationship between the quadratic equation and its roots helps us quadratic equation and its roots helps us save timesave time
Derivation of FormulaDerivation of Formulaaxax22 + bx + c = 0 has roots + bx + c = 0 has roots αα and and ββ –(1) –(1)
This tells us that:This tells us that: (x – (x – αα)(x – )(x – ββ) = 0) = 0
Expanding, we have:Expanding, we have: xx22 – – ααx – x – ββx + x + αβαβ = 0 = 0
xx22 –– ( (αα + + ββ)x + )x + αβαβ = 0 = 0
So, if we compare the above with (1):So, if we compare the above with (1):
xx22 + x + = 0 + x + = 0b
a
c
a
Result: Result: αβαβ = , = , αα + + ββ = = c
ab
a
In other words…In other words…
axax22 + bx + c = 0 has roots + bx + c = 0 has roots αα and and ββ
xx22 –– ( (αα + + ββ))xx + + αβαβ = 0 = 0
xx22 –– (sum of roots) (sum of roots)xx + product of roots = 0 + product of roots = 0
Application – Example 1 (TB Pg 55)Application – Example 1 (TB Pg 55)
2x2x22 + 6x – 3 = 0 has roots + 6x – 3 = 0 has roots αα and and ββ, find the value of, find the value of
(a) (a) 1 1
Without finding Without finding αα and and ββ, we know the value for , we know the value for αα + + ββ and and αβαβ
αα + + ββ = – 3 and = – 3 and αβαβ = – 3/2 = – 3/2
1 1 32
32
Hence,Hence,
Application – Example 1 (TB Pg 55)Application – Example 1 (TB Pg 55)
(b) (b) (2 1)(2 1)
Expanding (b), we haveExpanding (b), we have
We already know that: We already know that: αα + + ββ = – 3 and = – 3 and αβαβ = – 3/2 = – 3/2
2x2x22 + 6x – 3 = 0 has roots + 6x – 3 = 0 has roots αα and and ββ, find the value of, find the value of
(2 1)(2 1) 4 2 2 1
4 2( ) 1
34( ) 2( 3) 1
211
Application – Example 2 (TB Pg 55)Application – Example 2 (TB Pg 55)
(a) (a) 2 2
To find (a), we use the algebraic propertyTo find (a), we use the algebraic property
Step 1: Step 1: αα + + ββ = 2 and = 2 and αβαβ = – 4 = – 4
xx22 – 2x – 4 = 0 has roots – 2x – 4 = 0 has roots αα and and ββ, find the value of, find the value of
2 2 2( ) 2
Rearranging:Rearranging: 2 2 2
2
( ) 2
(2) 2( 4)
12
Application – Example 2 (TB Pg 55)Application – Example 2 (TB Pg 55)
(b) (b)
To find (b), we use the algebraic propertyTo find (b), we use the algebraic property
Step 1: Step 1: αα + + ββ = 2 and = 2 and αβαβ = – 4 = – 4
xx22 – 2x – 4 = 0 has roots – 2x – 4 = 0 has roots αα and and ββ, find the value of, find the value of
2 2 2( ) 2
Rearranging:Rearranging:2 2 2( ) 2
12 2( 4)
20
Hence, Hence, αα − − ββ = = 20
Application – Example 2 (TB Pg 55)Application – Example 2 (TB Pg 55)
(c) (c) 4 4
To find (a), we use the algebraic propertyTo find (a), we use the algebraic property
Step 1: Step 1: αα + + ββ = 2 and = 2 and αβαβ = – 4 = – 4
xx22 – 2x – 4 = 0 has roots – 2x – 4 = 0 has roots αα and and ββ, find the value of, find the value of
2 2 2 4 2 2 4( ) 2
Rearranging:Rearranging: 4 4 4 2 2
2 2
( ) 2
(12) 2( 4)
112
Application – Example 3 (TB Pg 56)Application – Example 3 (TB Pg 56)2x2x22 + 4x – 1 = 0 has roots + 4x – 1 = 0 has roots αα and and ββ, form the equation with roots , form the equation with roots αα22 and and ββ22
xx22 –– (sum of roots) (sum of roots)xx + product of roots = 0 + product of roots = 0
Remember!Remember!
xx22 –– ( (αα22 + + ββ22))xx + + αα22ββ22 = 0 = 0
So in order to get the equation, I must find So in order to get the equation, I must find αα22 + + ββ22 and and αα22ββ22
How? How?