Question

(1) x2 + 4kx + (k2-k + 2) = 0

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Answer 1

Answer:

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Step-by step explanation:

= x²+4kx +(k²-k+2) = 0

= From above equation.

A = 1 , B = 4k , C= k^2 - k +2

Now,

°.° It has equal roots .

.°. B^2 - 4AC =0

Equating the values.

= (4k)^2 -4 (1)(k^2 - k + 2) =0

=> 16k^2 - 4k^2 +4k -8 =0

=> 12k^2 + 4k -8 =0

=>4 [3k^2 + k -2 ] =0

=> 3k^2 +k -2 =0

=> 3k^2 +3k -2k -2 = 0

=> 3k (k +1) -2 (k+1) =0

=> (3k-2)(k+1) =0

=> 3k-2=0 OR k+1 = 0

=> k = 2/3 OR k = -1

So, the value of 'k' is 2/3 0r -1.

Answer 2

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