Question: (a) By substituting $p = 2^x$, show that the equation
$2 \times 4^x - 2^{x+3} = 17 \times 2^{x-1} - 4$
can be written in the form
$4p^2 - 33p + 8 = 0$
(b) Hence solve
$2 \times 4^x - 2^{x+3} = 17 \times 2^{x-1} - 4$
1. **State the problem:**
We want to show that by substituting $p = 2^x$, the equation
$$2 \times 4^x - 2^{x+3} = 17 \times 2^{x-1} - 4$$
can be rewritten as
$$4p^2 - 33p + 8 = 0$$
and then solve the original equation.
2. **Rewrite terms using $p = 2^x$:**
Recall that $4^x = (2^2)^x = 2^{2x} = (2^x)^2 = p^2$.
So,
$$2 \times 4^x = 2 \times p^2$$
and
$$2^{x+3} = 2^x \times 2^3 = p \times 8 = 8p$$
Similarly,
$$17 \times 2^{x-1} = 17 \times \frac{2^x}{2} = 17 \times \frac{p}{2} = \frac{17p}{2}$$
3. **Rewrite the equation substituting these expressions:**
$$2p^2 - 8p = \frac{17p}{2} - 4$$
4. **Multiply both sides by 2 to clear the fraction:**
$$2 \times (2p^2 - 8p) = 2 \times \left( \frac{17p}{2} - 4 \right)$$
which gives
$$4p^2 - 16p = 17p - 8$$
5. **Bring all terms to one side:**
$$4p^2 - 16p - 17p + 8 = 0$$
Simplify the $p$ terms:
$$4p^2 - 33p + 8 = 0$$
This matches the required form.
6. **Solve the quadratic equation:**
$$4p^2 - 33p + 8 = 0$$
Use the quadratic formula:
$$p = \frac{33 \pm \sqrt{(-33)^2 - 4 \times 4 \times 8}}{2 \times 4}$$
Calculate the discriminant:
$$(-33)^2 - 4 \times 4 \times 8 = 1089 - 128 = 961$$
Square root:
$$\sqrt{961} = 31$$
So,
$$p = \frac{33 \pm 31}{8}$$
Two solutions:
- $$p = \frac{33 + 31}{8} = \frac{64}{8} = 8$$
- $$p = \frac{33 - 31}{8} = \frac{2}{8} = \frac{1}{4}$$
7. **Recall $p = 2^x$, solve for $x$:**
- For $p = 8$:
$$2^x = 8 = 2^3 \implies x = 3$$
- For $p = \frac{1}{4}$:
$$2^x = \frac{1}{4} = 2^{-2} \implies x = -2$$
**Final answer:**
$$x = 3 \text{ or } x = -2$$