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

In our equation, the coefficients are:

• $a = 2k - 3$
• $b = -k$
• $c = k - 1$. Therefore, we substitute these coefficients into the discriminant: $(-k)^2 - 4(2k - 3)(k - 1) = 0.$ This simplifies to: $k^2 - 4(2k - 3)(k - 1) = 0.$

Next, we expand the equation: $k^2 - 4((2k^2 - 2k - 3k + 3)) = 0$ Simplifying gives: $k^2 - 4(2k^2 - 5k + 3) = 0$ Which further simplifies to: $k^2 - 8k^2 + 20k - 12 = 0$ So we have: $-7k^2 + 20k - 12 = 0$.

Rearranging gives us: $7k^2 - 20k + 12 = 0$. Next, we can use the quadratic formula to find the values of $k$: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 7$, $b = -20$, and $c = 12$: $k = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 7 \cdot 12}}{2 \cdot 7}$. After calculating the discriminant: $k = \frac{20 \pm \sqrt{400 - 336}}{14} = \frac{20 \pm \sqrt{64}}{14} = \frac{20 \pm 8}{14}.$ Thus, we obtain:

1. $k = \frac{28}{14} = 2$
2. $k = \frac{12}{14} = \frac{6}{7}$. Hence, the values of $k$ for which the equation has equal roots are $k = 2$ and $k = \frac{6}{7}$.