Question

The values of k for which the quadratic equation kx2 +1 = kx + 3x −11x2 has real and equal roots are (a)

Answer 1

Given

p (x ) = kx^2 + 1 = kx + 3x - 11x^2

has equal root

To find

k = ?

Solution

kx^2 + 1 = kx + 3x - 11x^2

kx^2 + 11x^2 - kx - 3x + 1 = 0

( k + 11 )x^2 - ( k + 3 )x + 1 = 0

a = ( k + 11 )

b = - ( k + 3 )

C = 1

b^2 - 4ac = 0

( - k - 3 )^2 - 4 ( k + 11 )( 1) = 0

k^2 + 6k + 9 - 4k - 44 = 0

k^2 + 2k - 35 = 0

k^2 + 7k - 5k - 35 = 0

k ( k + 7 ) - 5 ( k + 7 ) =

( k + 7 ) ( k - 5 ) = 0

k + 7 = 0. or. k - 5 = 0

k = - 7. or. k = 5

Answer 2

Answer:

If kx2+1=3x+kx−6x2 has real and equal roots, then k=