Stealth Quadratics – A Level Maths Revision

To solve stealth quadratics (or “disguised quadratics”), you begin by making a substitution. This substitution will turn the stealth quadratic into a normal-looking quadratic, which you can then factorise and solve.


Example Question 1

Find all real solutions to the equation π‘₯4 – 9π‘₯2 + 20 = 0

The hardest part about this question is spotting that it is actually a stealth quadratic equation (the more you practise these, the easier it will be to spot them). This means you need to find a substitution to turn it into a normal-looking quadratic equation.

Notice that the middle term is π‘₯2 and the first term, π‘₯4, equals (π‘₯2)2, so substitute 𝑦 = π‘₯2.

The equation becomes:

(π‘₯2)2 – 9(π‘₯2) + 20 = 0
𝑦2 – 9𝑦 + 20 = 0

Remember, at A Level you are encouraged to use your calculator to solve quadratic equations:

𝑦 = 5, 𝑦 = 4

Next, replace the substitution and solve these equations for π‘₯:

𝑦 = π‘₯2
π‘₯2 = 5, π‘₯2 = 4
π‘₯ = ±√5, π‘₯ = Β±2

Notice that all solutions to this equation are real. In other words, they belong to the set of real numbers. An equation that would have no real solutions would be, for example, one which involved finding the square root of a negative number.

Sometimes, the substitution you need to make will have a fractional power or be a trigonometric or exponential term.


Example Question 2

Solve the equation sin2(ΞΈ) + 2sin(ΞΈ) – 3 = 0 for 0Β° ≀ ΞΈ ≀ 180Β°

Notice that, if you substitute 𝑦 = sin(ΞΈ), the equation will look like a quadratic. The equation becomes:

[sin(ΞΈ)]2 + 2[sin(ΞΈ)] – 3 = 0
𝑦2 + 2𝑦 – 3 = 0
𝑦 = 1, 𝑦 = -3

Then, replace the substitution and solve those equations for ΞΈ:
𝑦 = sin(ΞΈ)
sin(ΞΈ) = 1, sin(ΞΈ) = -3

The only equation that has a solution is sin(ΞΈ) = 1, so you can disregard the second equation (since -1 ≀ sin(ΞΈ) ≀ 1).

ΞΈ = sin-1(1) = 90Β°
There are no other solutions in the interval 0Β° ≀ ΞΈ ≀ 180Β°

In very rare cases, you can use the substitution technique to solve inequalities. when doing this, you will need to be extremely careful to consider the shape of the graph before deciding which are the correct intervals.


Example Question 3

Use a suitable substitute to solve the inequality π‘₯4 + 3π‘₯2 – 4 > 0.

When solving non-linear inequalities (specifically, quadratic inequalities), begin by setting the expression equal to zero and solving the resulting equation. Then, you can identify the shape of the graph to help establish which parts of the function are negative (the parts of graph that lie below the π‘₯-axis) and which are positive (the parts of the graph that lie above the π‘₯-axis).

To solve π‘₯4 + 3π‘₯2 – 4 = 0, you will first need to make it look like a quadratic. Substitute 𝑦 = π‘₯2. The equation becomes:

𝑦2 + 3𝑦 – 4 = 0
𝑦 = -4, 𝑦 = 1

Since the equation π‘₯2 = -4 has no real solutions, solve π‘₯2 = 1 to find π‘₯ = Β±1.

These values of π‘₯ will provide the boundaries to the solutions to the inequality. To properly define the range where π‘₯4 + 3π‘₯2 – 4 is greater than 0, you need to sketch the graph.

It is a quartic graph (it’s highest power of π‘₯ is 4), so you know its shape. You can see the 𝑦-intercept is -4, and the π‘₯-intercepts (the roots we have found above) are 1 and -1. This means you can assume the graph looks like this:

You do not need to calculate any other turning points because you know there are only two π‘₯-intercepts, so the graph cannot cross the π‘₯-axis again. This means that any turning points between π‘₯ = -1 and π‘₯ = 1 cannot affect the result.

The graph lies above the π‘₯-axis for values of π‘₯ < -1 and > 1, and below for values of π‘₯ such that -1 < π‘₯ < 1. This means the range of solutions to the inequality π‘₯4 + 3π‘₯2 – 4 > 0 are π‘₯ < -1 and π‘₯ > 1.


Practice Questions

By using a suitable substitution, find all real solutions to the following equations.

a. x^4 - 2x^2 - 15 = 0

b. x^6 = 8x^3 - 16

c. (x - 2)^2 + 6(x - 2) - 72 = 0

d. x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 10 = 0

e. 3x^3 - 7x^{\frac{3}{2}} + 2 = 0 (give your answers in index form)

f. 2\cos^2(\theta) + 13\cos(\theta) - 7 = 0 (solutions in the range 0Β° ≀ ΞΈ ≀ 90Β°)

g. 6x^4 + 12x^2 = x2 + 7 (give your answers in index form)

h. 3x - 13\sqrt{x} + 4 = 0

i. x^2 + 40x^{-2} = 13

j. -2x^4-x^2+3 \le 0


Answers

a. x^4 - 2x^2 - 15 = 0

\boldsymbol{\begin{aligned} &\textbf{Let }y = x^2\\&y^2-2y-15=0\\&y=5,y=-3 \\&x^2=5,x^2=-3\\&\textbf{No real solutions to the equation }x^2=-3\\&x=\pm \sqrt{5} \end{aligned}}

b. x^6 = 8x^3 - 16

\boldsymbol{\begin{aligned} &x^6-8x^3+16=0\\&\textbf{Let } y=x^3\\&y^2-8y+16=0\\&y=4\\&x^2=4\\&x=\pm 2\end{aligned}}

c. (x - 2)^2 + 6(x - 2) - 72 = 0

\boldsymbol{\begin{aligned}&\textbf{Let }y=x-2\\&y^2+6y-72=0\\&y=-12,y=6\\&x-2=-12, x-2 = 6\\&x=-10, x=8 \end{aligned}}

d. x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 10 = 0

\boldsymbol{\begin{aligned} &\textbf{Let} y=x^{\frac{1}{3}}\\&y^2+3y-10=0\\&y=2, y=-5\\&x^{\frac{1}{3}}=2, x^{\frac{1}{3}}=-5,\\&x=8, x=-125\end{aligned}}

e. 3x^3 - 7x^{\frac{3}{2}} + 2 = 0 (give your answers in index form)

\boldsymbol{\begin{aligned}&\textbf{Let }y=x^{\frac{3}{2}}\\&3y^2-7y+2=0\\&y=\frac{1}{3}, y=2\\&x^{\frac{3}{2}}=\frac{1}{3}, x^{\frac{3}{2}}=2\\&x=3^{-\frac{2}{3}}, x=2^{\frac{2}{3}}, \textbf{or equivalent}\end{aligned}}

f. 2\cos^2(\theta) + 13\cos(\theta) - 7 = 0 (solutions in the range 0Β° ≀ ΞΈ ≀ 90Β°)

\boldsymbol{\begin{aligned} &\textbf{Let }y=\cos(\theta)\\&2y^2+13y-7=0\\&y=\frac{1}{2}, y=-7\\&\cos(\theta)=\frac{1}{2}, \cos(\theta)=-7\\&\textbf{No real solutions to the equation }\cos(\theta)=-7\\&\theta=\cos^{-1}(\frac{1}{2})=60^{\circ} \end{aligned}}

g. 6x^4 + 12x^2 = x2 + 7 (give your answers in index form)

\boldsymbol{\begin{aligned} &6x^4+11x^2-7=0\\&\textbf{Let }y=x^2\\&6y^2+11y-7=0\\&y=-\frac{7}{3},y=\frac{1}{2}\\&x^2=-\frac{7}{3}, x^2=\frac{1}{2}\\&\textbf{No real solutions to the equation }x^2=-\frac{7}{3}\\&x=\pm\sqrt{\frac{1}{2}}=\pm 2^{-\frac{1}{2}}\end{aligned}}

h. 3x - 13\sqrt{x} + 4 = 0

\boldsymbol{\begin{aligned} &\textbf{Let }y=\sqrt{x}\\&3y^2-13y+4=0\\&y=\frac{1}{3}, y=4\\&\sqrt{x}=\frac{1}{3}, \sqrt{x}=4\\&x=\frac{1}{9}, x=16\end{aligned}}

i. x^2 + 40x^{-2} = 13

\boldsymbol{\begin{aligned} &\textbf{Multiply through by }x^2\\&x^4-13x^2+40=0\\[1em]&\textbf{Let }y=x^2\\&y^2-13y+40=0\\&y = 8, y=  5\\&x^2=8, x^2=5\\&x=\pm \sqrt{8}=\pm 2\sqrt{2}, x=\pm\sqrt{5}\end{aligned}}

j. -2x^4-x^2+3 \le 0

\boldsymbol{\begin{aligned} &\textbf{Let }y=x^2 \\&-2y^2-y+3 \le 0\\[1em]&\textbf{Solutions to }-2y^2-y+3=0 \textbf{ are:}\\&y=1, y=-\frac{3}{2}\\&x^2=1,x^2=-\frac{3}{2}\\[1em]&\textbf{The only equation with real solutions is: }\\&x^2=1, x=\pm 1\\[1em]&\textbf{As }-2x^4-x^2+3\textbf{ has a negative leading coefficient,}\\&\textbf{the graph is inverted and looks like this:}\end{aligned}}

\boldsymbol{\begin{aligned} &\textbf{As the function is negative when the graph} \\&\textbf{lies below the x-axis, the solutions are:}\\&x \le -1, x \ge 1\end{aligned}}


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