Solving Higher Order Quadratics by Substitution
Some higher order equations are quadratics in disguise. In \(x^4-13x^2+36=0\), the substitution \(u=x^2\) turns the equation into \(u^2-13u+36=0\), which factorises, and each value of \(u\) then gives values of \(x\). In NSW Year 10 Advanced maths the substitution can be a power, a bracket, a root, a reciprocal or an exponential (NESA outcome MA5‑EQU‑P‑02).
You will solve quartic and sixth-power equations, equations in a repeated bracket, and equations using \(u=\dfrac{1}{x}\), \(u=\sqrt{x}\), \(u=2^x\) and \(u=x^{1/3}\). You will reject values of the substitution that give no real solutions, find a missing constant from one known solution, and find the sides of a rectangle from its area and diagonal. It is part of Year 10 Advanced Quadratics.
Every question with a fully worked solution.
- Solving Higher Order Quadratics by Substitution - Video - Solving Equations In Quadratic Form Using Substitution Watch
Theory
An equation such as \(x^4-17x^2+16=0\) is a quadratic in disguise. Let \(u=x^2\), solve \(u^2-17u+16=0\), then change back to \(x\). This Year 10 Advanced maths page shows how to spot the substitution, how to change back, and how to reject values such as \(x^2=-6\) or \(\sqrt{x}=-2\) that have no real solution.
An equation is reducible to a quadratic when it has the shape \(a(\ldots)^2+b(\ldots)+c=0\), where the same expression fills both gaps. Replacing that expression by a single letter, usually \(u\), is called a substitution.
Common substitutions: \(u=x^2\) for \(x^4\) and \(x^2\); \(u=x^3\) for \(x^6\) and \(x^3\); \(u=\sqrt{x}\) for \(x\) and \(\sqrt{x}\); \(u=\dfrac{1}{x}\) for \(\dfrac{1}{x^2}\) and \(\dfrac{1}{x}\); \(u=2^x\) for \(4^x=2^{2x}\) and \(2^x\).
After solving for \(u\), change back to \(x\). Each value of \(u\) can give two, one or no values of \(x\): \(x^2=9\) gives \(x=\pm 3\), \(x^3=8\) gives \(x=2\) only, and \(x^2=-4\) gives nothing.
Some values of \(u\) are impossible: a square \(x^2\) is never negative, \(\sqrt{x}\) is never negative, and a power such as \(2^x\) is always positive. Reject them and say why.
Pick \(u\) so the equation becomes \(au^2+bu+c=0\):
| Equation contains | Let | Becomes | Change back with |
|---|---|---|---|
| \(x^4\) and \(x^2\) | \(u=x^2\) | \(au^2+bu+c=0\) | \(x=\pm\sqrt{u}\), needs \(u\ge 0\) |
| \(x^6\) and \(x^3\) | \(u=x^3\) | \(au^2+bu+c=0\) | \(x=\sqrt[3]{u}\), one value |
| \(x\) and \(\sqrt{x}\) | \(u=\sqrt{x}\) | \(au^2+bu+c=0\) | \(x=u^2\), needs \(u\ge 0\) |
| \(\dfrac{1}{x^2}\) and \(\dfrac{1}{x}\) | \(u=\dfrac{1}{x}\) | \(au^2+bu+c=0\) | \(x=\dfrac{1}{u}\), needs \(u\neq 0\) |
| \(x^{2/3}\) and \(x^{1/3}\) | \(u=x^{1/3}\) | \(au^2+bu+c=0\) | \(x=u^3\) |
| \(k^{2x}\) and \(k^x\) | \(u=k^x\) | \(au^2+bu+c=0\) | write \(u\) as a power of \(k\), needs \(u\gt 0\) |
| a bracket and its square | \(u=\) the bracket | \(au^2+bu+c=0\) | solve bracket \(=u\) |
How to solve by substitution
- Rearrange so one side is zero, and write the terms in order of power.
- Spot the expression whose square also appears. Let \(u\) equal it.
- Rewrite the equation in \(u\) and solve it by factorising, the quadratic formula or completing the square.
- Change back: set the expression equal to each value of \(u\) and solve for \(x\).
- Reject impossible cases (a negative square, a negative square root, a non-positive power) and give the reason.
- Check at least one answer in the original equation.
Other question types
- Show there are no real solutions: find every value of \(u\) and explain why each one is impossible, such as \(x^2=-1\) and \(x^2=-3\).
- A given solution: substitute it to find an unknown constant first, then solve the full equation by substitution.
- Products: \(x^2(x^2-5)=36\) expands to \(x^4-5x^2-36=0\), which suits \(u=x^2\).
- Geometry: a rectangle with area \(A\) and diagonal \(d\) gives \(w=\dfrac{A}{l}\) and \(l^2+\dfrac{A^2}{l^2}=d^2\). Multiplying by \(l^2\) gives an equation in \(l^4\) and \(l^2\). Keep only positive lengths.
(i) \(x^4-17x^2+16=0\)
(ii) \(x^4+2x^2-24=0\)
(i) Let \(u=x^2\), so \(x^4=u^2\).
| \(u^2-17u+16\) | \(=\) | \(0\) |
| \((u-1)(u-16)\) | \(=\) | \(0\) |
| \(x^2=1 \text{ or } x^2\) | \(=\) | \(16\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm 1} \text{ or } \boldsymbol{x=\pm 4}\) |
Four solutions: \(-4\), \(-1\), \(1\) and \(4\).
(ii) Again let \(u=x^2\).
| \(u^2+2u-24\) | \(=\) | \(0\) |
| \((u+6)(u-4)\) | \(=\) | \(0\) |
| \(x^2=-6 \text{ or } x^2\) | \(=\) | \(4\) |
\(x^2=-6\) has no real solution, because a square is never negative. So \(\boldsymbol{x=\pm 2}\) only.
(i) \(x^6+26x^3-27=0\)
(ii) \((x^2+x)^2-8(x^2+x)+12=0\)
(i) Let \(u=x^3\), so \(x^6=u^2\).
| \(u^2+26u-27\) | \(=\) | \(0\) |
| \((u+27)(u-1)\) | \(=\) | \(0\) |
| \(x^3=-27 \text{ or } x^3\) | \(=\) | \(1\) |
| \(x\) | \(=\) | \(\boldsymbol{-3} \text{ or } \boldsymbol{x=1}\) |
A cube root has only one real value, and it can be negative: \((-3)^3=-27\).
(ii) Let \(u=x^2+x\).
| \(u^2-8u+12\) | \(=\) | \(0\) |
| \((u-2)(u-6)\) | \(=\) | \(0\) |
| \(x^2+x-2=0\) | \(\Rightarrow\) | \((x+2)(x-1)=0\) |
| \(x^2+x-6=0\) | \(\Rightarrow\) | \((x+3)(x-2)=0\) |
\(\boldsymbol{x=-3,\ -2,\ 1}\) or \(\boldsymbol{2}\).
(i) \(x-\sqrt{x}-6=0\)
(ii) \(\dfrac{2}{x^2}+\dfrac{1}{x}-1=0\)
(i) \(x=(\sqrt{x})^2\), so let \(u=\sqrt{x}\).
| \(u^2-u-6\) | \(=\) | \(0\) |
| \((u-3)(u+2)\) | \(=\) | \(0\) |
| \(\sqrt{x}=3 \text{ or } \sqrt{x}\) | \(=\) | \(-2\) |
\(\sqrt{x}\) means the positive square root, so \(\sqrt{x}=-2\) is impossible. Then \(\sqrt{x}=3\) gives \(\boldsymbol{x=9}\). Check: \(9-3-6=0\).
(ii) \(\dfrac{2}{x^2}=2\left(\dfrac{1}{x}\right)^2\), so let \(u=\dfrac{1}{x}\).
| \(2u^2+u-1\) | \(=\) | \(0\) |
| \((2u-1)(u+1)\) | \(=\) | \(0\) |
| \(\dfrac{1}{x}=\dfrac{1}{2} \text{ or } \dfrac{1}{x}\) | \(=\) | \(-1\) |
| \(x\) | \(=\) | \(\boldsymbol{2} \text{ or } \boldsymbol{x=-1}\) |
(i) \(3^{2x}-10\times 3^x+9=0\)
(ii) \(x^2-2x+\dfrac{8}{x^2-2x}=6\), exactly
(i) \(3^{2x}=(3^x)^2\), so let \(u=3^x\).
| \(u^2-10u+9\) | \(=\) | \(0\) |
| \((u-1)(u-9)\) | \(=\) | \(0\) |
| \(3^x=1=3^0 \text{ or } 3^x\) | \(=\) | \(9=3^2\) |
| \(x\) | \(=\) | \(\boldsymbol{0} \text{ or } \boldsymbol{x=2}\) |
(ii) Let \(u=x^2-2x\). Then \(u+\dfrac{8}{u}=6\). Multiply by \(u\):
| \(u^2-6u+8\) | \(=\) | \(0\) |
| \((u-2)(u-4)\) | \(=\) | \(0\) |
\(x^2-2x-2=0\) and \(x^2-2x-4=0\) do not factorise, so use the quadratic formula on each.
| \(x\) | \(=\) | \(\dfrac{2\pm\sqrt{12}}{2}=1\pm\sqrt{3}\) |
| \(x\) | \(=\) | \(\dfrac{2\pm\sqrt{20}}{2}=1\pm\sqrt{5}\) |
\(\boldsymbol{x=1\pm\sqrt{3}}\) or \(\boldsymbol{x=1\pm\sqrt{5}}\). None of them makes \(x^2-2x=0\), so all four are solutions.
Common pitfalls
Frequently asked questions
What is a substitution in algebra?
It means replacing an expression with a single letter to make the equation simpler. In \(x^4-7x^2+12=0\), letting \(u=x^2\) gives the quadratic \(u^2-7u+12=0\).
How do you know which substitution to use?
Look for one term that is the square of another: \(x^4\) and \(x^2\), \(x\) and \(\sqrt{x}\), \(4^x\) and \(2^x\), or a bracket and its square. Let \(u\) equal the smaller one.
How many solutions can an equation in x to the fourth have?
Up to four real solutions, because each of the two values of \(u\) can give two values of \(x\). It has fewer when a value of \(u\) is negative or zero.
Why do some values of u get rejected?
Because \(u\) stands for something with a restricted range. \(x^2\) and \(\sqrt{x}\) cannot be negative, and \(2^x\) is always positive, so any value of \(u\) outside that range gives no real \(x\).
Do I have to check my answers?
It is good practice, and it is essential with square roots and fractions, where a step such as squaring or multiplying by an expression can create values that do not satisfy the original equation.