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Question
Find the values of k for which the roots are real and equal in each of the following equation:
\[4 x^2 - 2\left( k + 1 \right)x + \left( k + 1 \right) = 0\]
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Solution
The given equation is \[4 x^2 - 2(k + 1)x + (k + 1) = 0\] where a = 4, b = -2(k+1), c = (k+1)
As we know that D = b2 - 4ac
Putting the value of a = 4, b = -2(k+1), c = (k+1)
\[\left\{ - 2(k + 1) \right\}^2 - 4 \times 4 \times (K + 1)\]
\[4(K + 1 )^2 - 16(K + 1)\]
\[(K + 1)\left\{ 4(K + 1) - 16 \right\}\]
\[(K + 1)(4K - 12)\]
\[4(K + 1)(K - 3)\]
For real and equal roots D = 0
\[4\left( K + 1 \right)\left( K - 3 \right) = 0\]
\[K = - 1 \text { or } k = 3\]
Therefore, the value of
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