Question

2kx2 - 4x + k = 0 what is the roots of it​

Answer 1

Let a be the common roots of the two given quadratic equations then, we must have;

a^2 - 4a + k = 0 = a^2 + ka - 4 . Now on substraction it ===> a = (k +4)/(k +4 ) = 1 ( provided k not equal to -4 ) , if k = - 4 , note that both the equations become same and they are x^2 - 4x - 4 = 0 or (x - 2 )^2 - 8 = 0 which in turn implies (x - 2 )^2 = 8 or x = 2 (+or -) 2 sqrt(2) and these are the required roots in that case.

Next,let k not equal to -4 , then a = 1 , when substituted in a^2 - 4a + k =0 gives k = 3 . Now for k = 3 , the first eq. becomes x^2 - 4x + 3 = (x - 1)( x - 3 )= 0 ; so that x = 1 , 3 . Similarly for k= 3 the second eq. becomes, x^2 + 3 x - 4 = ( x - 1 )(x + 4 ) = 0 giving x = 1 , - 4 . This way 1 becomes the common root and 3 , -4 are their second but different roots .