# What two numbers has sum equal to 9/4 and the product equal to 3/4?

Jan 7, 2017

$\frac{9}{8} + \frac{\sqrt{33}}{8} \text{ }$ and $\text{ } \frac{9}{8} - \frac{\sqrt{33}}{8}$

#### Explanation:

Notice that given two numbers $a$ and $b$:

$\left(x - a\right) \left(x - b\right) = {x}^{2} - \left(a + b\right) x + a b$

So the two numbers we want will be the zeros of the quadratic polynomial:

$f \left(x\right) = {x}^{2} - \frac{9}{4} x + \frac{3}{4}$

So we find:

$0 = 64 f \left(x\right)$

$\textcolor{w h i t e}{0} = 64 \left({x}^{2} - \frac{9}{4} x + \frac{3}{4}\right)$

$\textcolor{w h i t e}{0} = 64 {x}^{2} - 144 x + 48$

$\textcolor{w h i t e}{0} = {\left(8 x\right)}^{2} - 2 \left(8 x\right) \left(9\right) + 81 - 33$

$\textcolor{w h i t e}{0} = {\left(8 x - 9\right)}^{2} - {\left(\sqrt{33}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(8 x - 9\right) - \sqrt{33}\right) \left(\left(8 x - 9\right) + \sqrt{33}\right)$

$\textcolor{w h i t e}{0} = \left(8 x - 9 - \sqrt{33}\right) \left(8 x - 9 + \sqrt{33}\right)$

Hence:

$x = \frac{1}{8} \left(9 \pm \sqrt{33}\right) = \frac{9}{8} \pm \frac{\sqrt{33}}{8}$

That is, the two numbers are:

$\frac{9}{8} + \frac{\sqrt{33}}{8} \text{ }$ and $\text{ } \frac{9}{8} - \frac{\sqrt{33}}{8}$