The roots of the following quadratic equation are real and equal then find the value of K.
(1) kx(x-2)+6=0
Answers
Answered by
16
Hey!!!!
__________
We have
=> kx² - 2kx + 6 = 0 with real and equal roots
Thus. D = 0
=> b² - 4ac = 0
Substituting values
=> (-2k)² - 4(k)(6) = 0
=> 4k² - 24k = 0
=> 4k(k - 6) = 0
Thus => k = 0 or k - 6 = 0
=> k = 6
_________
Hope this helps ✌️
Good Morning
__________
We have
=> kx² - 2kx + 6 = 0 with real and equal roots
Thus. D = 0
=> b² - 4ac = 0
Substituting values
=> (-2k)² - 4(k)(6) = 0
=> 4k² - 24k = 0
=> 4k(k - 6) = 0
Thus => k = 0 or k - 6 = 0
=> k = 6
_________
Hope this helps ✌️
Good Morning
Khushi54811:
A very very good and beautiful morning and very much thank you for answer
welcome
Answered by
15
HELLO DEAR,
GIVEN THAT:-
kx² - 2kx + 6 = 0 with real and equal roots
Where
a = k
b = -(2k)
c = 6
∴ D = 0
=> b² - 4ac = 0
NOW, put the values
⇒ (-2k)² - 4(k)(6) = 0
⇒ 4k² - 24k = 0
⇒ 4k(k - 6) = 0
⇒ k = 0 or k = 6
I HOPE ITS HELP YOU DEAR,
THANKS
GIVEN THAT:-
kx² - 2kx + 6 = 0 with real and equal roots
Where
a = k
b = -(2k)
c = 6
∴ D = 0
=> b² - 4ac = 0
NOW, put the values
⇒ (-2k)² - 4(k)(6) = 0
⇒ 4k² - 24k = 0
⇒ 4k(k - 6) = 0
⇒ k = 0 or k = 6
I HOPE ITS HELP YOU DEAR,
THANKS
thank you so so much for the help
:-)
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