# The sum of two numbers is 6 and their product is 4. How do you find the larger of the two numbers?

Jul 1, 2015

Write the conditions as two equations and solve to obtain:
the larger of the two numbers is $3 + \sqrt{5}$

#### Explanation:

Let the two numbers be $x$ and $y$

We are told that
[1]$\textcolor{w h i t e}{\text{XXXX}}$$x + y = 6$
and
[2]$\textcolor{w h i t e}{\text{XXXX}}$$x y = 4$

Rearranging [1] we have
[3]$\textcolor{w h i t e}{\text{XXXX}}$$y = 6 - x$

Substituting [3] into [2]
[4]$\textcolor{w h i t e}{\text{XXXX}}$$x \left(6 - x\right) = 4$

Which simplifies as
[5]$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 6 x + 4 = 0$

Using the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

[6]$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{6 \pm \sqrt{36 - 16}}{2}$

[7]$\textcolor{w h i t e}{\text{XXXX}}$$x = 3 \pm \sqrt{5}$

Since in [1] and [2] $x$ and $y$ are symmetric, they share the same solution possibilities.

The larger of these possibilities is $3 + \sqrt{5}$

May 9, 2016

Write an equation and solve it.

The larger number is 5.236..

#### Explanation:

It is possible to do this using one variable.
If two numbers add up to 6, they can be written as $x \mathmr{and} \left(6 - x\right)$

Their product is 4 $\Rightarrow x \left(6 - x\right) = 4$

$6 x - {x}^{2} = 4 \text{ "rArr x^2 - 6x + 4 = 0" a quadratic}$

This does not factorise, but it is a good example for using completing the square because $a = 1 \mathmr{and} \text{b is even}$

${x}^{2} - 6 x + \text{ " = -4 " + move the constant}$

${x}^{2} - 6 x + \text{??? " = -4 " + ???}$
${x}^{2} - 6 x + 9 \text{ " = -4 + 9" }$add ${\left(\frac{b}{2}\right)}^{2} \text{to both sides}$
${\left(x - 3\right)}^{2} = 5$
$x - 3 = \pm \sqrt{5}$

$x = 3 + \sqrt{5} = 5.236 \text{ } \mathmr{and} x = 3 - \sqrt{5} = 0.764$

5.236 is the larger.