If a and ß are the roots
of the
quadratic equation3x^2-6x+1=0 then find the quadratic
equation whose
roots are
2a+ ß and 2ß+a
pls answer this and I will mark ur answer as brainliest
Answers
If α and β are the roots of the equation 3 x² - 6 x + 1 =0,form an equation whose roots are
(i) 1/α , 1/β
(ii) α² β , β² α
(iii) 2 α + β , 2 β + α
SOLUTION :
I had taken a = α and b = β
GIVEN :
3 x² - 6 x + 1 =0
By comparing the given equation with general form of quadratic equation ax² + bx + c = 0
Here, a = 3 b = - 6 and c = 1
Sum of the roots (α + β )= -b/a
α + β = - (-6)/3
α + β= 2…………..(1)
Product of roots (α β) = c/a
α β= ⅓…………….(2)
Here α = 1/α and β = 1/β. [GIVEN]
Sum of the roots= (α + β )
α + β= 1/α + 1/β
α + β = (α + β)/αβ
α + β = 2/(⅓) = 2×3 = 6 [ from eq 1 &2 ]
α + β = 6 ………(3)
Product of roots = (α β)
α β= (1/α) (1/β)
α β = 1/αβ
α β = 1/(⅓) = 3. [ from eq 2 ]
α β = 3…………….(4)
General form of quadratic equation whose roots are α and β
x² - (sum of roots) x + product of roots
x² - (α + β) x + α β = 0
x² - 6 x + 3= 0
Hence, the required quadratic equation is x² - 6 x + 3= 0
ii)
By comparing the given equation with general form of quadratic equation ax² + bx + c = 0 get a = 3 b = -6 and c = 1
Here, a = 3 b = - 6 and c = 1
Sum of the roots (α + β )= -b/a
α + β = - (-6)/3
α + β= 2…………..(1)
Product of roots (α β) = c/a
α β= ⅓…………….(2)
Here α = α² β and β = β² α
Sum of roots = α + β
= (α² β + β² α)
α + β= α β (α + β)
α + β = (1/3) (2)= ⅔ [ from question 1 and 2]
α + β = ⅔…………(3)
Product of roots= (α β)
= (α² β) (β² α)
= α³ β³
α β = (α β)³
α β = (1/3)³ = 1/27 [ from eq 2]
α β = 1/27………….(4)
General form of quadratic equation whose roots are α and β
x² - (sum of roots) x + product of roots
x² - (α + β) x + α β = 0
x² - (2 /3) x+ (1/27) = 0
[From eq 3 and 4]
(27 x² - 18 x + 1)/27 = 0
27 x² - 18 x + 1 = 0
Hence, the required quadratic equation is 27 x² - 18 x + 1 = 0
(iii) 2 α + β , 2 β + α
Here α = 2 α + β and β = 2 β + α
Sum of roots = α + β
= (2 α + β + 2 β + α) = 3 α + 3 β
α + β= 3 (α + β)
α + β = 3(2) [ from eq 1 of ques 1]
α + β = 6 …………(1)
Product of roots= (α β)
= (2 α + β) (2 β + α)
= 4 α β + 2 α² + 2 β² + α β)
α β = 5 α β + 2(α² + β²)
[α² + β² = (α + β)² - 2 α β]
α β = 5 α β + 2(α + β)² - 2 α β]
[ From eq 1 & 2 of Question 1]
α β = 5 × ⅓ + 2 (2)² - 2× ⅓
α β = 5/3 + 2 × (4 -⅔ )
α β = 5/3 + 2 (12-2)/3
α β = 5/3 + 2 ×10/3
α β = 5/3 + 20/3= 25/3
α β = 25/3…….(2)
General form of quadratic equation whose roots are α and β
x² - (sum of roots) x + product of roots
x² - (α + β) x + α β = 0
x² - 6x + 25/3 = 0
(3 x² - 18 x + 25)/3 = 0
3 x² - 18 x + 25 = 0
Hence, the required quadratic equation is 3 x² - 18 x + 25 = 0.