# Symmetrically So

Exploit the symmetry and turn this quartic into a quadratic.

## Problem

Make a substitution to find two exact real solutions to the equation $(x + 3)^4 + (x + 5) ^4 = 20.$

Did you know ... ?

Frequently mathematicians spend their time stuck wondering how to solve equations or problems. One way of cracking a tough problem is to make a transformation to turn it into a more familiar form which allows the solution to proceed. Finding good substitutions or transformations is one of the more creative aspects of mathematics.

Frequently mathematicians spend their time stuck wondering how to solve equations or problems. One way of cracking a tough problem is to make a transformation to turn it into a more familiar form which allows the solution to proceed. Finding good substitutions or transformations is one of the more creative aspects of mathematics.

## Getting Started

Try to solve this equation without multiplying it out. Notice a symmetry and make a substitution to a new variable and you can quickly solve this quartic by turning it into a quadratic...

If you are trying to find the square root of $4+2\sqrt{3}$ (i.e. you are trying to find $x$ such that $x^2=4+2\sqrt{3}$) then you can write $x=a+b\sqrt{3}$ and then find $a$ and $b$ such that $(a+b\sqrt{3})^2=4+2\sqrt{3}$.

## Student Solutions

In this case make the substitution $y=x+4$, which transforms the equation into

$$

(y-1)^4+(y+1)^4=20.

$$

This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.

If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:

$$

(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.

$$

So, the odd factors cancel to give

$$

2y^4+12y^2-18=0

$$

This then becomes the simple equation

$$

y^4+6y^2-9=0\,,

$$

which is a quadratic equation in $y^2$ with solutions

$$

y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).

$$

One of these solutions is positive; taking the square root gives two real values for $y$ as

$$

y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}

$$

Therefore two real solutions to the equation are

$$

x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}

$$