if x=2 and x=3 are roots of the equation 3x2-2kx +2m = 0,find the value of k and m
Answers
Answered by
436
Solution :
_____________________________________________________________
Given :
2 & 3 are the roots of the equation, 3x² - 2kx + 2m = 0.
_____________________________________________________________
To Find :
The value of k & m
_____________________________________________________________
If x = 2,
We get,
⇒ 3x² - 2kx + 2m = 0
⇒ 3(2)² - 2k(2) + 2m = 0
⇒ 3(4) - 4k + 2m = 0.
⇒ 12 - 4k + 2m = 0
⇒ -4k + 2m = -12
⇒ 2k - m = 6 ....(i)
__________________
If x = 3,
Then,
⇒ 3x² - 2kx + 2m = 0
⇒ 3(3)² - 2k(3) + 2m = 0
⇒ 3(9) - 6k + 2m = 0
⇒ 27 - 6k + 2m = 0
⇒ -6k + 2m = -27
⇒ 3k - m = 13. 5..(ii)
____________________
Subtracting equation (i) from (ii),
We get,
⇒ (3k - m) - (2k - m) = 13.5 - 6
⇒ 3k - m - 2k + m = 7.5
⇒ ∴ k = 7.5
_______________________
Substituting value of x in (i),
We get,
⇒ 2k - m = 6
⇒ 2(7.5) - m = 6
⇒ 15 - m = 6
⇒ -m = 6 - 15
⇒ -m = -9
⇒ ∴ m = 9
_____________________________________________________________
Hope it Helps !!
_____________________________________________________________
Given :
2 & 3 are the roots of the equation, 3x² - 2kx + 2m = 0.
_____________________________________________________________
To Find :
The value of k & m
_____________________________________________________________
If x = 2,
We get,
⇒ 3x² - 2kx + 2m = 0
⇒ 3(2)² - 2k(2) + 2m = 0
⇒ 3(4) - 4k + 2m = 0.
⇒ 12 - 4k + 2m = 0
⇒ -4k + 2m = -12
⇒ 2k - m = 6 ....(i)
__________________
If x = 3,
Then,
⇒ 3x² - 2kx + 2m = 0
⇒ 3(3)² - 2k(3) + 2m = 0
⇒ 3(9) - 6k + 2m = 0
⇒ 27 - 6k + 2m = 0
⇒ -6k + 2m = -27
⇒ 3k - m = 13. 5..(ii)
____________________
Subtracting equation (i) from (ii),
We get,
⇒ (3k - m) - (2k - m) = 13.5 - 6
⇒ 3k - m - 2k + m = 7.5
⇒ ∴ k = 7.5
_______________________
Substituting value of x in (i),
We get,
⇒ 2k - m = 6
⇒ 2(7.5) - m = 6
⇒ 15 - m = 6
⇒ -m = 6 - 15
⇒ -m = -9
⇒ ∴ m = 9
_____________________________________________________________
Hope it Helps !!
Answered by
122
Hence the ans is k =15/2 where as m=9
Attachments:
Similar questions