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JEE Advanced2009MathematicsQuadratic EquationActual

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

Correct answer

2

Step-by-step solution

(i) Given, x^2-8 k x+16 (k^2-k+1 )=0 Now, D=64 k^2- (k^2-k+1 ) aligned & =64(k-1)>0 & k>1 & 16-32 k+16 (k^2-k+1 ) 0 & k^2-3 k+2 0 & (k-2)(k-1) 0 k 1 or k 2 & aligned Hence, k=2

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