Find the values of k for which the equation (2k-3)x^2-kx+(k-1) has equal roots

We know that an equation has equal roots if the sqrt(b^2-4ac) term in the quadratic equation is equal to zero. Therefore using this information we can form an expression for k to be (-k)^2-4(2k-3)(k-1)=0. From this we can simplfy the expression by expanding the brackets to give -7k^2+20k-12=0, which is in the form of a quadratic equation which we can solve to find two values of k.

Knowing that 7 is a prime number we know that one bracket has to contain -7k and the other to contain k. We then look at the factors of -12 and knowing that we need to make a large value of 20k realise that the minus interger term has to be in the bracket with k. Looking at the factors of -12 it becomes obvious that using -2 and 6 will yield 20k and therefore we are able to find the equation can be simplfied to (-7k+6)(k-2)=0 giving k=2,6/7 are the values for k