Question

(iii) k2 x2 – 2(2k – 1)x + 4 = 0​

Answer 1

Question :

Find the value of k for which the quadratic equation k²x² - 2(2k - 1)x + 4 = 0 has real and equal roots .

Answer :

k = 1/4

Solution :

Here ,

The given quadratic equation is ;

k²x² - 2(2k - 1)x + 4 = 0 .

Comparing the given quadratic equation with the general quadratic equation ax² + bx + c = 0 , we have ;

a = k²

b = -2(2k - 1)

c = 4

For real and equal roots , it's discriminant must be zero .

Thus ,

=> D = 0

=> b² - 4ac = 0

=> [-2(2k - 1)]² - 4•k²•4 = 0

=> 4(4k² + 1 - 4k) - 4•4k² = 0

=> 4(4k² + 1 - 4k - 4k²) = 0

=> 1 - 4k = 0

=> 4k = 1

=> k = 1/4

Hence , k = 1/4 .