If the roots of x3−9x2+kx+l=0 are in A.P with common difference 2 then (k, I)=
Answers
Step-by-step explanation:
Given:-
The roots of x^3−9x^2+kx+l=0 are in A.P with common difference 2
To find:-
Find: (k, I) ?
Solution:-
Given cubic equation is x^3−9x^2+kx+l=0---(1)
On Comparing this with the standard cubic equation ax^3+bx^2+cx+d = 0
a = 1
b = -9
c= k
d = l
Given that the roots are in the AP
Common difference = 2
Let the roots be (a-2) ,a ,(a+2)
Sum of the roots = -b/a
=> a-2+a+a+2 = -(-9)/1
=> 3a = 9
=> a = 9/3
=> a = 3
d = 2
Then the three roots are 3-2,3,3+2
=> 1,3,5
Sum of the roots = 1+3+5 = 9
Product of the roots = (1)(3)(5)=15
Sum of the product of roots taken at a time
= (1)(3)+(3)(5)+(5)(1)
= 3+15+5
= 23
We know that
If α β and γ are the roots then the cubic equation is x^3-(α +β +γ)x^2+(α β+βγ+γα)x-α β γ=0
=> x^3-9x^2+23x-15 = 0------(2)
On comparing both (1) &(2)
k = 23
l = -15
(k,l) = (23,-15)
Answer:-
The value of (k,l) for the given problem is (23,-15)
Used formulae:-
1.the standard cubic equation is ax^3+bx^2+cx+d = 0
2.If α β and γ are the roots then the cubic equation is x^3-(α +β +γ)x^2+(α β+βγ+γα)x -αβγ=0
- Sum of the roots =-b/a
- Sum of the product of roots taken at a time=c/a
- Product of the roots =-d/a