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  • The smallest integral value of k , for which both the root of the equation x^2-8kx+16(k^2-k+1)=0are real ,distinct and have value at least 4 is

The smallest integral value of k , for which both the root of the equation x^2-8kx+16(k^2-k+1)=0are real ,distinct and have value at least 4 is

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Solution: As x^2-8kx+16(k^2-k+1)=0 has real and distinct roots .

A

(-8k)^2-(4)(16)(k^2-k+1)> 0Rightarrow k-1> 0 As k is an integer, we get kgeq 2.

If alpha ,eta are roots of given equation ,then alpha ,eta geq 4

\Rightarrow alpha +eta geq 8;and;alpha eta geq 16\ \Rightarrow 8kgeq 8;and;16(k^2-k+1)geq 16\ \Rightarrow kgeq 1;and;k(k-1)geq =0\ \Rightarrow kgeq 1

Thus, smallest value of k is 2

Posted by

Deependra Verma

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