Given y = 2^x; show that 2^{x+1} - 17(2^x) + 8 = 0 can be written in the form 2y^2 - 17y + 8 = 0 (b) Hence solve 2^{x+1} - 17(2^x) + 8 = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1

Given the equation:

$2y^2 - 17y + 8 = 0$

1. We can use the quadratic formula, where: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, a = 2, b = -17, and c = 8.

2. We calculate the discriminant: $b^2 - 4ac = (-17)^2 - 4 \cdot 2 \cdot 8 = 289 - 64 = 225$

3. Thus: $y = \frac{17 \pm \sqrt{225}}{2 \cdot 2} = \frac{17 \pm 15}{4}$ This gives us two possible values for y:

   • Case 1: $y = \frac{32}{4} = 8$
   • Case 2: $y = \frac{2}{4} = \frac{1}{2}$

4. Now, we resolve for x using the original substitution y = 2^x:

   • For y = 8: $2^x = 8 \Rightarrow x = 3$
   • For y = \frac{1}{2}: $2^x = \frac{1}{2} \Rightarrow x = -1$

5. The solution to the equation thus provides the values: $x = 3\quad \text{or} \quad x = -1$