Find the values of $k$ for which the equation $(2k - 3)x^2 - kx + (k - 1) = 0$ has equal roots. - AQA - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

In our equation, we identify the coefficients:

• $a = (2k - 3)$
• $b = -k$
• $c = (k - 1)$.
  Thus, setting the discriminant gives us:
  $(-k)^2 - 4(2k - 3)(k - 1) = 0.$

Now we simplify the expression:
$k^2 - 4((2k - 3)(k - 1)) = 0.$
Expanding this, we find: k^2 - 4[(2k^2 - 2k - 3k + 3)] = 0\ This leads to:
k^2 - 4(2k^2 - 5k + 3) = 0\
k^2 - (8k^2 - 20k + 12) = 0\
$(7k^2 - 20k + 12 = 0).$

We now solve the quadratic equation $7k^2 - 20k + 12 = 0$ using the quadratic formula:
$k = \frac{-b \\pm \sqrt{b^2 - 4ac}}{2a} = \frac{20 \\pm \sqrt{(-20)^2 - 4(7)(12)}}{2(7)}$
Calculating the discriminant, we find:
400 - 336 = 64.\
So we have:
$k = \frac{20 \\pm 8}{14}.$
This gives us two solutions:
k = \frac{28}{14} = 2\
and
$k = \frac{12}{14} = \frac{6}{7}.$
Thus, $k$ can be either $2$ or $\frac{6}{7}$.