Question

Check whether the roots of the quadratic equation 1 /x-2 + 2/x-1 =6/x are real or not. If yes find the roots (x not equal to 0, 1,2)​

Answer 1

Step-by-step explanation:

1/(x-2) + 2/(x-1) = 6/x

(x-1) + 2(x-2) /(x-1)(x-2) = 6/x

x - 1 + 2x -4 / x^2 -2x -x + 2 = 6/x

(3x -5)/ x^2 -3x +2 = 6/x

x ( 3x-5) = 6 ( x^2 - 3x +2)

3x^2 -5x = 6x^2 - 18x + 12

3x^2 - 13x + 12 = 0

to find the nature's of the roots we have to find determiner.

D = b^2 -4ac

= (13)^2 -4 × 3 × 12

= 169 - 144

D = 25

hence D>0 so the equation has equal roots.

3x^2. - 13x + 12= 0

3x^2 -(9+4)x+12 =0

3x^2 -9x-4x +12=0

3x ( x -3) -4(x-3)=0

(x-3) (3x-4) =0

so

x= 3, 4/3