If 2 and 3 are zeroes of polynomial 3 x2 - 2kx + 2m, then find the values of k and m. with full method
Answers
Step-by-step explanation:
here a=3,b=2k,c=2m
sum of Zeroes =2+3=5
product of Zeroes =2×3=6
5=-b/a;6=c/a
5=-2k/3;6=2m/3
k =15/-2;m =18/2
:k=-15/2;m=9 mark as bralinest plz
Step-by-step explanation:
Given :-
2 and 3 are zeroes of a polynomial
3x²- 2kx + 2m
To find :-
Find the values of k and m ?
Solution:-
Method-1:-
Given quadratic polynomial=3x²-2kx+2m
Let P(x) = 3x²- 2kx + 2m
On comparing with the standard quadratic polynomial ax²+bx+c
We have ,
a = 3
b = -2k
c = 2m
Given zeroes of P(x) = 2 and 3
We know that
Sum of the zeroes = -b/a
=> 2+3 = -(-2k)/3
=> 5 = 2k/3
=> 5×3 = 2k
=> 15 = 2k
=> 2k = 15
=> k = 15/2
and
Product of the zeroes = c/a
=> 2×3 = 2m/3
=> 6 = 2m/3
=>6×3 = 2m
=> 18 = 2m
=> 2m = 18
=> m = 18/2
=> m = 9
Therefore, k = 15/2 amd m = 9
Method -2:-
Given quadratic polynomial=3x²-2kx+2m
Given zeroes of P(x) = 2 and 3
We know that
The Quadratic Polynomial whose zeroes are α and β is K[x²-(α +β)x+α β]
=> K[x²-(2+3)x+(2×3)]
=> K[x²-5x+6]
If K = 1 then the required polynomial is x²-5x+6
Now , Since 2 and 3 are the zeroes of 3x²-2kx+2m then
=> 3x²- 2kx + 2m = x²-5x+6
=> x²-(2k/3)x+(2/3)m = x²-5x+6
On comparing both sides then
=> 2k/3 = 5 and 2m/3 = 6
On taking 2k/3 = 5
=> 2k = 3×5
=> 2k = 15
=> k = 15/2
and
2m/3 = 6
=>2m = 6×3
=> 2m = 18
=> m = 18/2
=> m = 9
Therefore, k = 15/2 amd m = 9
Answer:-
The value of k = 15/2 and m = 9
Used formulae:-
- The Quadratic Polynomial whose zeroes are α and β is K[x²-(α +β)x+α β]
- The standard quadratic Polynomial ax²+bx+c
- Sum of the Zeroes = -b/a
- Product of the zeroes = c/a