The equation whose roots are -2, 3±✓5 is
Answers
Step-by-step explanation:
Given :-
The roots are -2, 3±✓5
To find :-
Write the equation ?
Solution :-
Method-1:-
Given roots are -2, 3±✓5
They are -2, 3+✓5, 3-✓5
Since given roots are 3 then the equation must be a Cubic equation.
Let α = -2
Let β = 3+√5
Let γ = 3-√5
I) Sum of the roots = α+ β+ γ
= -2+3+√5+3-√5
= 6-2
= 4
II)Sum of the product of two roots taken at a time = αβ +βγ +γα
= (-2)(3+√5)+(3+√5)(3-√5)+(3-√5)(-2)
=> -6-2√5+9-5-6+2√5
=> -17+9
=> -8
and
III) Product of the roots = α β γ
= (-2)(3+√5)+3-√5)
=> (-2)(3²-(√5)²)
=> (-2)(9-5)
=> (-2)(4)
=> -8
We know that
The equation whose roots are α, β , γ is
x³-(α+β +γ)x²+(αβ +βγ +γα)x-α β γ = 0
=> x³-(4)x²+(-8)x-(-8) = 0
=> x³-4x²-8x+8 = 0
Method -2:-
Given roots are -2, 3±✓5
They are -2, 3+✓5, 3-✓5
Since given roots are 3 then the equation must be a Cubic equation.
We know that
If -2 is the root then (x+2) is a factor
If (3+√5) is the root then (x-(3+√5)) is a factor
If (3-√5) is the root then (x-(3-√5)) is a factor.
Therefore,
The equation whose roots are α, β ,γ is
(x-α)(x-β)(x- γ) = 0
The required equation is
(x+2)(x-(3+√5))(x-(3-√5)) = 0
=> (x+2)([(x-3-√5)(x-3+√5)] = 0
=> (x+2)([(x-3)²-(√5)²] = 0
=> (x+2)(x²-6x+9-5) = 0
=> (x+2)(x²-6x+4) = 0
=> x(x²-6x+4)+2(x²-6x+4) = 0
=> x³-6x²+4x+2x²-12x+8 = 0
=> x³-4x²-8x+8 = 0
Answer :-
The required equation whose roots are -2, 3±√5 is x³-4x²-8x+8 = 0
Used formulae:-
»»The equation whose roots are α, β , γ is
x³-(α+β +γ)x²+(αβ +βγ +γα)x-α β γ = 0
»» The equation whose roots are α, β ,γ is (x-α)(x-β)(x- γ) = 0
»»The Cubic equation has at most three roots.