Question

Find the roots of the quadratic equation
1/(x-3) – 1/(x-6) = 9/20; x ≠ 3, 6

Answer 1

Took LCM of the terms in left hand side

LCM will be (x-3)(x-6)

So in the denominator of left hand side it will be x^2 -9x +18
While in the numerator the equation will be (x-6)-(x-3)

Which will be simplified to -3

Now the equation will be -3 / (x^2 -9x + 18) = 9/ 20

Now divide both the side by 3 and cross multiply the equation , you will get
-20= 3x^2 - 27x +54

Now add 20 to both the sides of equation , you will get..

0=3x^2 - 27x +74

Now use the quadratic formula that if a equation is in the form ax^2 + bx + c =0
Then x = [-b +- root over( b^2 - 4ac)]/2a

Substitute the values ..

The values of x will be imagining... you can check

The value of x will be [27 +- root over (-159) ] / 6

Hope it was helpful to you