RRB Group D 3-Oct-2018 Shift 3 Solved Paper

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Question : 44 of 100

Marks: +1, -0

If

6 {x}^{2}+2 {kx}+{k}=0

has no real solution, then the value of k will satisfy :

$K>-6$
$k>6$
$0<{k}<6$
$k<6$

Solution:

Given equation

6 x^{2}+2 k x+k=0

For no real solution,

{D}<0

(2 k)^{2}-6 ×4 ×k<0

k^{2}-6 k<0

k(k-6)<0

$∴0

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