# Agile Algebra

Observe symmetries and engage the power of substitution to solve complicated equations.

## Problem

Equations can be difficult to solve by direct attack but if you look for symmetric features and make substitutions they might become much easier to solve.

Consider the equation:$$(x+5)^2+(x+3)^2=6$$

**Expand the brackets and solve the resulting quadratic equation, leaving your answers in exact form**

The algebra here can get a little messy. Perhaps we can use a substitution to turn the equation into an easier one.

**Use the substitution $y=x+4$ to rewrite the equation in terms of $y$.***Why do you think that we have suggested using $y=x+4$? How does this substitution relate to the original equation?*

If $y=x+4$ then $x+5=y+1$ and $x+3=y-1$. This means that we can write the original equation as $(y+1)^2+(y-1)^2=6$.

**Expand your brackets and solve for $y$, and then for $x$.**

**Which method was easiest?**

Try using substitutions to simplify and then solve these equations:

1. $(x + 3)^4 + (x + 5) ^4 = 34$

2. $(x-1)(x-3)(x-5)(x-7)+15=0$

3. $(x-4)(x-5)(x-6)(x-7)=1680$

The "median" of these brackets is equal to $x-5\frac 1 2$, so try using a substitution of $t=x-5\frac 1 2$. You might find that a further substitution is helpful.

Now lets consider a different equation: $$x^4-8x^3 + 17x^2-8x+1=0$$

There is symmetry in the coefficients of the equation. Dividing by $x^2$ gives: $$x^2 -8x+17 -\dfrac 8 x +\dfrac 1{x^2}=0$$

Consider a substitution of $z=x+\dfrac 1 x$. What will $z^2$ be?

If $z=x+\dfrac 1 x$ then $z^2=\left(x+\dfrac {1} {x}\right)^2=x^2 + 2 + \dfrac 1 {x^2}$

Can you write $x^2 -8x+17 -\dfrac 8 x +\dfrac 1{x^2}=0$ in terms of $z$ and $z^2$?

You can rearrange the terms to get $x^2 +17 +\dfrac 1{x^2} - 8x - \dfrac 8 x=0$.

You should find that you have a quadratic in $z$ which you can solve to get two solutions. For each solutions for $z$, substitute it into $z=x+\dfrac 1 x$ and solve for $x$.

Try using substitutions to simplify and then solve these equations:

4. $x^4-2x^3 -6x^2-2x+1=0$

5. $x^4-2x^3+2x^2-2x+1=0$

**Extension**

Try using a substitution to help solve this equation. There is a hint available!

$$(8x+7)^2(4x+3)(x+1)=\frac{9}{2}$$

You can multiply the $(4x+3)$ bracket by $2$ and the $(x+1)$ bracket by $8$ - this will mean that the left hand side of the equation has multiplied by $16$, so do the same to the right hand side!

NOTES AND BACKGROUND

This is an example of a process which occurs frequently in mathematics. Let's refer to two frames of reference as A and B and say we have a problem stated in A, then the technique is to map the given relationships to B, work in B and then map the results back to A. All these equations have symmetry of one sort or another. By using the symmetry to make a substitution you can change the variable
and get an equation which is easier to solve. After that you have to use the solutions you have found and go back to find the corresponding solutions of the original equation.

To find out more about this general technique see the article "The Why and How of Substitution".

## Student Solutions

1. $(x + 3)^4 + (x + 5) ^4 = 34$

Nayanika from the Tiffin Girls' School in the UK solved this equation neatly, using a substitution. There is one small error which is corrected.

*The solution is $x=-4\pm\sqrt2.$*

2. $(x-1)(x-3)(x-5)(x-7)+15=0$

Nayanika used another substitution:

*Remember that $y=-2,-\sqrt6$ are also solutions, which give $x=2, 4-\sqrt6$*

3. $(x-4)(x-5)(x-6)(x-7)=1680$

Patrick from Woodbridge shool sent in this solution:

Let us rewrite the equation with $a = x-7$, then we have $(a+3)(a+2)(a+1)a = 1680.$

Note that $a(a+3) = a^2 + 3a$, and $(a+2)(a+1) = a^2 + 3a + 2$, so $(a+2)(a+1)(a+3)a = (a^2 + 3a)(a^2 + 3a + 2).$ Let us then take $y = a^2 + 3a$, then $y(y+2) = 1680$ and $y^2 + 2y - 1680 = 0.$

Factorising, we find $(y-40)(y+42) = 0$, so $y = -42$ or $+40$. Since $y = a^2 + 3a$, then we have either: $40 = a^2 + 3a$, so $(a+8)(a-5) = 0$ and $a = -8$ or $5$.

Alternatively $-42 = a^2 + 3a$, which has no real roots. Therefore, we have $a = -8$ or $5$. Then, since $a = x - 7$, $x = a + 7$, so $x = -1$ or $12$.

We will test this by substitution: $(-1-4)(-1-5)(-1-6)(-1-7)$, so for each bracket (-a-b) we can take out the negative to give -(a+b). There are four such brackets, so there are four -1 factors removed, which multiply to give 1. Therefore, $(1+4)(1+5)(1+6)(1+7) = 5\times 6\times 7\times 8 = 1680.$ Similarly, $x=12$ gives us $8\times 7\times 6\times 5 = 1680.$

Well done Patrick. An alternative substitution could be $t= x - 5\frac{1}{2}.$

Tom from The Skinners School used a substitution for the equation $x^4-8x^3 + 17x^2-8x+1=0,$ but it is not the one we suggested. Here is Tom's solution:

$f(x) = x^4 - 8x^3 + 17x^2 - 8x + 1 = 0$

use substitution $z = x^2 - 8x + 1$

so $f(x) = zx^2 + 15x^2 +z = 0$

use same substitution again for the $x^2$ term that is multiplied with $z$

$f(x) = z(z^2 + 8x - 1) + 15x^2 + z = 0$

$= z^2 + 8xz - z + 15x^2 + z$

$= z^2 + 8xz + 15x^2$

$= (z+3x)(z+5x)$

so $x^2 - 5x + 1 = 0$

and $x^2 - 3x +1 =0$

you can solve these quadratics using the quadratic equation and this gives you the four solutions to $f(x) = 0.$

Could the substitution we suggested, $z = x + \frac1x$ help with the next two equations?

4. $x^4-2x^3 -6x^2-2x+1=0$

5. $x^4-2x^3+2x^2-2x+1=0$

Tom also sent in a solution to the extension, $(8x+7)^2(4x+3)(x+1)=\frac{9}{2}$

let $y = x+1$

$(8y-1)^2(4y-1)(y) = \frac{9}{2}$

$(64y^2-16y+1)(4y^2-y) =\frac{9}{2}$

let $z = 4y^2 - y$

$(16z+1)(z) = \frac{9}{2}$

$16z^2 + z =\frac{9}{2}$

$32z^2 + 2z - 9 = 0$

$(2z-1)(16z+9) = 0$

Second bracket gives no real values of $y$ so for real solutions, $2z - 1 =0$

$z = \frac{1}{2}$

$4y^2 - y = \frac{1}{2}$

$8y^2 -2y -1=0$

$(4y+1)(2y-1)=0$

$y= -\frac{1}{4} \text{and} \frac{1}{2}$

$x = -\frac{5}{4} \text{and} -\frac{1}{2}$

## Teachers' Resources

**Why do this problem?**

Making substitutions to make the task in hand easier is an example of a valuable technique in many areas of mathematics (eg integration, transformations of functions, change of axes, diagonalisation of matrices etc.). By introducing students to this technique as an example of a general process it will help them to understand what is going on when they meet the process in many different mathematical situations. These equations give students useful practice in algebraic manipulation. They will need to look for symmmetrical features in the expressions and exploit the symmetry to make it easier to solve the equations.

**Possible approach**

*This problem featured in the **NRICH Secondary webinar** in June 2022.*

It is helpful to introduce this problem with some discussion about switching frames of reference to make the equations easier to solve and how you are looking for symmetry in order to choose a good substitution.

Start by asking how pupils might solve $(x+3)^2+(x+5)^2=6$.

Follow up by asking if they notice anything about the expressions in the brackets. Can they relate them to another expression?

Introduce the idea of a substitution, such as $y=x+4$. Ask how each of the brackets is related to $x+4$ and how we could rewrite the equation in terms of $y$. Ask pupils to then solve the equation in terms if $y$. How can we then find $x$?

Ask students which method they though was easiest - then ask them to solve $(x+3)^4+(x+5)^4=34$.

There are several other equations in this problem where symmetry can be used to find a helpful substitution.

**Possible extension ideas**

- $(2x - 3)^4 + (2x - 5)^4 = 2$
- $ (x - 2)(x + 1)(x + 4)(x + 7) = 19$
- $(12x - 1)(6x - 1)(4x - 1)(3x - 1) = 5$

**Possible support**

Make up your own equation by taking a simple equation and making a substitution to make it more complicated. For example take any quadratic equation in $x$ and turn it into a quartic equation by substituting $x=z+\frac{1}{z}.$