Find the value of k for which the eguation has real roots - kx(x-2)= -6
Anonymous:
whose Ans is correct?
yours
but by mistake i marked the other as brainliest
then y u select his ans as brainlist?
it's ok.
sorry bro
it's okay... when u select his ans as brainlist i thought that u think mine procedure is wrong.. that's y i ask you
Answers
Answered by
0
Hey
b^2 - 4ac = 0
(2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
4k^2 = 24k
k = 6
no this is not correct
since b^2-4ac can also be>0
Answered by
8
Heya!
1) ist possibility
ROOTS ARE REAL AND DISTINCT
-kx² + 2kx + 6 = 0
FOR REAL AND DISTINCT ROOTS DESCRIMNANT MUST BE > 0
ITS DESCRIMNANT IS GIVEN BY
( 2k )² - 4 ( - k ) ( 6 ) >0
4k² + 24k > 0
4k ( k + 6 ) >0
k > 0 OR k > -6
k € ( 0 , -6 )
2) 2nd possibility
ROOTS ARE REAL AND EQUAL
In this case Descrimnant must be = 0
it's Descrimnant is given by
( 2k )² + 24k = 0
4k² + 24k = 0
k = 0 OR k = -6
that's correct ☺☺☺
hey friend can you tell me whats this- € as you used it in answer
€ means belongs to
whats belongs to
what it means
it means k can take all values between ( 0 to - 6 )
ok thanks
and when do we use this symbol
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