Find the values of k for which the equation (2k-3)x^2- kx+(k-1)=0 has equal roots.

For a quadratic to have equal roots, the value of the discriminant (b²-4ac) =0. This means that the quadratic formula yields only one result (or two equal roots), since x=(-b ±0)/2a , thus if a and b are constants, the 'equal' values of x are the constant -b/2a. This gives us a new quadratic. If a=2k-3, b=k and c=k-1, (b²-4ac) =0=k²-4(2k-3)(k-1).Expanding the brackets we get k²-8k²+8k+12k-12=0 or -7k²+20k-12=0. If we use the quadratic formula k=(-20±√(400-4(-7)(-12)))/2(-7) =(-20±√(400-336)/(-14)=(-20±8)/-14. Thus we get our two answers for k as k=-28/-14 or k=-12/-14, in their simplest form k=2, k=6/7.
To check your answer, you simply have to put it back into the first equation:For k=2:(2(2)-3)x² - (2)x +(2) -1 = 0 ---> x² -2x +1 =0 Factorise: (x-1)(x-1) thus the equal roots are x=1
For k=6/7:(2(6/7)-3)x² - (6/7)x +(6/7)-1=0 ----> (-9/7)x² - (6/7)x - (1/7)=0 or 9x² +6x + 1=0 Factorise: (3x+1)(3x+1) thus the equal roots are x=-1/3