**Solve for x: 2 ^{2x} – 8(2^{x}) + 15 = 0. Give your answer correct to two decimal places.**

This is easier than might first appear if we realise that it is a quadratic equation in terms of 2^{x}.

2^{2x} – 8(2^{x}) + 15 = 0

(2^{x})^{2} – 8(2^{x}) + 15 = 0

So if we let y = 2^{x}, then (2^{x})^{2} – 8(2^{x}) + 15 = 0 can be rewritten as y^{2} – 8y + 15 = 0.

We can use factorisation or our “minus b” formula to solve for y.

Using factorisation: (y-5)(y-3) = 0

y = 5 or y = 3

But since y = 2^{x}, then 2^{x} = 5, or 2^{x} = 3.

To solve when x is the exponent we use from the rules of indices and logarithms the fact that

a^{x} = y <=> log_{a}y = x

2^{x} = 5 <=> log_{2}5 = x = 1.61 to two decimal places

2^{x} = 3 <=> log_{2}3 = x = 1.10 to two decimal places

## Vertex Form of a Quadratic Equation

The vertex form of a quadratic equation is given by **y = a(x – h)**^{2}** + k**, where (h, k) is the vertex of the parabola.

The h represents the horizontal shift. How far left or right the graph is shifted from x = 0 from the parent equation y = x^{2}.

And k represents the vertical shift. How far up or down the graph is shifted from y = 0 from the parent equation y = x^{2}.

**Example: Sketch a graph of y = 5 – 3(x+2) ^{2}.**

Comparing **y = 5 – 3(x+2)**^{2}** **with** ****y = **a(x – h)^{2} + k, we get (h,k) = (-2,5) for the vertex of the parabola.

**To Convert from Vertex Form to ****y**** = ****ax**^{2}** + ****bx + c**** Form:**

Simply multiply out and combine like terms:

**y = 5 – 3(x+2)**^{2}** **

= 5 – 3(x^{2}+4x+4)

= 5- 3x^{2}-12x-12

= – 3x^{2}-12x-7

Using minus B formula, roots are: **x = -3.291** or **x = -0.709**