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The smallest value of k, for which both ...

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

Text Solution

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The correct Answer is:
`k=2`
(i) Given, `x^(2)-8kx+1(k^(2)-k+1)=0`
`Now, D=64{ k^(2)-k+1}=64 (k-1)gt0 k gt 1`
`(ii)-(b)/(2a)gt4implies(8k)/(2)gt4impliesk gt1`
`(iii) f(4)ge0`
`implies16-32k+16(k^(2)-k+1)ge0`
`impliesk^(2)-3k+2ge0`
`implies(k-2)(k-1)ge0`
`impliesk le1 ork ge2`
Hence, `k=3`
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