Question 1 :
If α and β are the roots of
2x2-3x-5 = 0
form a quadratic equation whose roots are α2 and β2
Solution :
By comparing the given equation with general form of quadratic equation we get,
a = 2, b = -3 and c = -5
Sum of the roots : α+β = -b/a α+β = 3/2 |
Product of roots : α β = c/a αβ = -5/2 |
here α = α2 and β = β2
General form of quadratic equation whose roots are α2 and β2
x2-(α2+β2)x+α2β2 = 0 ---(1)
x2-(α2+β2) x+(αβ)2 = 0
α2+β2 = (α+β)2-2αβ
= (3/2)2 - 2 (-5/2)
= (9/4) + 5
α2+β2 = 29/4
By applying the values in (1), we get
x2-(29/4)x+(-5/2)2 = 0
4x2-29x+25 = 0
Therefore the required quadratic equation is
4x2-29x+25 = 0
Question 2 :
If α and β are the roots of
x2-3x+2 = 0
form a quadratic equation whose roots are -α and -β.
Solution :
By comparing the given equation with general form of quadratic equation we get,
a = 1, b = -3 and c = 2
Sum of the roots : α+β = -b/a α+β = 3 |
Product of roots : αβ = c/a αβ = 2 |
Here α = -α and β = -β
General form of quadratic equation whose roots are α2 and β2
x2-(-α-β)x+(-α)(-β) = 0
x2+(α+β)x+αβ = 0 -----(1)
By applying the appropriate values in (1), we get
x2-3x+2 = 0
Therefore the required quadratic equations is
x2-3x+2 = 0
Question 3 :
If α and β are the roots of
x2-3x-1 = 0
then form a quadratic equation whose roots are 1/α2 and 1/β2.
Solution :
x2-3x-1 = 0
a = 1, b = -3 and c = -1
Sum of the roots : α+β = -b/a = 3 |
Product of roots : αβ = c/a = -1 |
Here α = 1/α2and β = 1/β2
By applying the above roots in the general form of a quadratic equation, we get
x2-(1/α2+1/β2)x+(1/α2) (1/β2) = 0
x2-(α2+β2/α2β2)x+[(1/α)(1/β)]2 = 0
x2-[(α2+β2)/(αβ)2]x+(1/αβ)2 = 0 ----(1)
α2+β2 = (α+β)2-2αβ
= 32-2(-1)
= 9+2
α2+β2 = 11
By applying the value of α2+β2 and αβ in (1), we get
x2-11x+1 = 0
So, the required quadratic equation is x2-11x+1 = 0.
Question 4 :
If α and β are the roots of the equation
3x2-6x+1 = 0
form an equation whose roots are
(i) 1/α , 1/β (ii) α2β , β2α (iii) 2α+β, 2β+α
Solution :
From the given quadratic equation
a = 3, b = -6 and c = 1
Sum of the roots : α+β = -b/a ==> 6/3 α+β = 2 |
Product of roots αβ = c/a αβ = 1/3 |
Here α = 1/α and β = 1/β
x2 - (1/α+1/β)x+(1/α) (1/β) = 0
x2-[(α + β)/αβ]x+[(1/αβ)] = 0 ----(1)
By applying the appropriate values in (1), we get
x2-6x+(2/3) = 0
3x2-18x+2 = 0
So, the required quadratic equation is
3x2-18x+2 = 0
(ii) α2β , β2α
Solution :
Here α = α2β and β = β2α
Applying the given roots instead of α and β in the general form.
x2-(α2β+β2α)x+(α2β)(β2α) = 0
x2-αβ(α+β)x+(α3β3) = 0
x2-αβ(α+β)x+(αβ)3 = 0 ----(1)
By applying the appropriate values in (1), we get
x2-(2x/3)+(1/27) = 0
27x2-18x+1 = 0
So, the required quadratic equation is
27x2-18x+1 = 0
(iii) 2α+β, 2β+α
Solution :
Here α = 2α+β and β = 2β+α
α+β = 2α+β+2β+α = 3α+3β α+β = 3(α+β) |
αβ = (2α+β)(2β+α) = 4αβ+2α²+2β²+αβ αβ = 5αβ+2(α2+β2) |
x2-3(α+β)x+5αβ+2(α2+β2) = 0 ----(1)
α2+β2 = (α+β)2-2αβ
= 22 - 2 (1/3)
= 4-2/3
α2+β2 = 10/3
By applying the values of α2+β2, α+β and αβ in (1), we get
x2-6x+(5/3)+(20/3) = 0
x2-6x+(25/3) = 0
3x2-18x+25 = 0
So, the required quadratic equation is
3x2-18x+25 = 0
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