(3k+1)x^2 + 2(k+1)x +1 find value …

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Sia ? 6 years, 6 months ago
The given equation is:
(3k + 1)x2 + 2(k + 1)x + 1 = 0
This is of the form ax2 + bx + c = 0, where
a = 3k + 1, b = 2(k + 1) = 2k + 2 and c = 1
As it is given that the given equation has real and equal roots, i.e., D = b2 - 4ac = 0.
{tex}\Rightarrow{/tex} (2k + 2)2 - 4(3k + 1) (1) = 0
{tex}\Rightarrow{/tex} 4k2 + 8k + 4 - 12k - 4 = 0
{tex}\Rightarrow{/tex} 4k2 - 4k = 0
{tex}\Rightarrow{/tex} 4k(k - 1) = 0
Therefore, either 4k = 0 or k - 1 = 0
{tex}\Rightarrow{/tex} k = 0 or k = 1
Hence, the roots of given equation are 1 and 0.
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