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 8 - 2017 - Paper 1

We have the quadratic equation from part (a):

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

To solve this, we will use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 2$, $b = -17$, and $c = 8$.

Calculating the discriminant:

$b^2 - 4ac = (-17)^2 - 4 \cdot 2 \cdot 8 = 289 - 64 = 225$

Next, we substitute back into the quadratic formula:

$y = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}$

Calculating the two possible solutions:

1. $y = \frac{32}{4} = 8$
2. $y = \frac{2}{4} = \frac{1}{2}$

Since we defined $y = 2^x$, we can now solve:

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

Therefore, the solutions are:

$x = 3, -1$