1.1.6 Hidden Quadratics - GCEALevelMaths9709

In this topic we will learn how to:

recognise and solve equations in $x$ which are quadratic in some function of $x$

A quadratic equation is one that can be written in the form $ax^{2} + bx + c$ . However, there are some functions that initially may not appear to satisfy this form but upon further inspection actually do satisfy it. These are known as quadratic equations in some function of $x$ or hidden quadratics.

Let’s look at some examples and how to solve them.

1. Solve the equation $x^{4} - 5x^{2} + 4 = 0$ .
$x^{4} - 5x^{2} + 4 = 0$ This does not look like a quadratic, however, we can rewrite it in the form $ax^{2} + bx + c$ to show that it is indeed a quadratic,
$(\textcolor{#2192ff}{x^{2}})^{2} - 5\textcolor{#2192ff}{x^{2}} + 4 = 0$ Let’s replace $\textcolor{#2192ff}{x^{2}}$ with $\textcolor{#2192ff}{y}$ ,
$\textmd{Let }\textcolor{#2192ff}{y} = \textcolor{#2192ff}{x^{2}}$ $(\textcolor{#2192ff}{y})^{2} - 5\textcolor{#2192ff}{y} + 4 = 0$ Now we can clearly see our quadratic in the form $ax^{2} + bx + c$ ,
$y^{2} - 5y + 4 = 0$ The next step is to solve the quadratic equation using your preferred method. I will use factorisation,
$(y - 4)(y - 1) = 0$ Equate each bracket to $0$ ,
$y - 4 = 0\ \ \ \ \ \ y - 1 = 0$ Solve the two linear equations,
$y = 4\ \ \ \ \ \ y = 1$ Earlier we said ‘let $y = x^{2}$ ‘. We are going to use that equation since we want to solve for $x$ ,
$\textcolor{#2192ff}{y} = x^{2}$ $\textmd{At } y = 4\ \ \ \ \ \ \textmd{At } y = 1$ $\textcolor{#2192ff}{4} = x^{2}\ \ \ \ \ \ \textcolor{#2192ff}{1} = x^{2}$ $\pm\sqrt{4} = \sqrt{x^{2}}\ \ \ \ \ \ \pm\sqrt{1} = \sqrt{x^{2}}$ $x = \pm 2\ \ \ \ \ \ x = \pm 1$ Therefore, our final answer is,
$x = \pm 1,\ \ x = \pm 2$ 2. Solve the equation $6x + \sqrt{x} - 1 = 0$ .
$6x + \textcolor{#2192ff}{\sqrt{x}} - 1 = 0$ Using laws of indices we can rewrite $\textcolor{#2192ff}{\sqrt{x}}$ as $\textcolor{#2192ff}{x^{\frac{1}{2}}}$ ,
$6x + \textcolor{#2192ff}{x^{\frac{1}{2}}} - 1 = 0$ This does not look like a quadratic, however, we can rewrite it to show that it is indeed a quadratic,
$6(\textcolor{#2192ff}{x^{\frac{1}{2}}})^{2} + \textcolor{#2192ff}{x^{\frac{1}{2}}} - 1 = 0$ Let’s replace $\textcolor{#2192ff}{x^{\frac{1}{2}}}$ with $\textcolor{#2192ff}{y}$ ,
$\textmd{Let } \textcolor{#2192ff}{y} = \textcolor{#2192ff}{x^{\frac{1}{2}}}$ $6(\textcolor{#2192ff}{y})^{2} + \textcolor{#2192ff}{y} - 1 = 0$ Now we can clearly see our quadratic in the form $ax^{2} + bx + c$ ,
$6y^{2} + y - 1 = 0$ The next step is to solve the quadratic equation using your preferred method. I will use factorisation,
$(3y - 1)(2y + 1) = 0$ Equate each bracket to $0$ ,
$3y - 1 = 0\ \ \ \ \ \ 2y + 1 = 0$ Solve the two linear equations,
$y = \frac{1}{3}\ \ \ \ \ \ y = -\frac{1}{2}$ Earlier we said ‘let $y = x^{\frac{1}{2}}$ ‘. We are going to use that equation since we want to solve for $x$ . When $\textcolor{#2192ff}{y}$ is $\textcolor{#2192ff}{\frac{1}{3}}$ ,
$\textcolor{#2192ff}{y} = x^{\frac{1}{2}}$ $\textcolor{#2192ff}{\frac{1}{3}} = x^{\frac{1}{2}}$ $\left(\frac{1}{3}\right)^{2} = \left(x^{\frac{1}{2}}\right)^{2}$ $x = \frac{1}{9}$ When $\textcolor{#0f0}{y}$ is $\textcolor{#0f0}{-\frac{1}{2}}$ ,
$\textcolor{#0f0}{y} = x^{\frac{1}{2}}$ $\textcolor{#0f0}{-\frac{1}{2}} = x^{\frac{1}{2}}$ $-\frac{1}{2} = \sqrt{x}$ $x = \textmd{no solutions}$ Note: The square root function does not produce negative numbers, therefore, we disregard that solution.

Therefore, our final answer is,
$x = \frac{1}{9}$ 3. Solve the equation $\tan^{2}{x} = 1 + \tan{x}$ .

Rearrange the equation so that all the terms are on one side,
$\tan^{2}{x} - \tan{x} - 1= 0$ This does not look like a quadratic, however, we can rewrite it to show that it is indeed a quadratic,
$(\textcolor{#2192ff}{\tan{x}})^{2} - \textcolor{#2192ff}{\tan{x}} - 1 = 0$ Let’s replace $\textcolor{#2192ff}{\tan{x}}$ with $\textcolor{#2192ff}{y}$ ,
$\textmd{Let } \textcolor{#2192ff}{y} = \textcolor{#2192ff}{\tan x}$ $(\textcolor{#2192ff}{y})^{2} - \textcolor{#2192ff}{y} - 1 = 0$ Now we can clearly see our quadratic in the form $ax^{2} + bx + c$ ,
$y^{2} - y - 1 = 0$ Since we cannot factorize by sight, we will use the quadratic formula,
$\textcolor{#2192ff}{a = 1}\textbf{, }\textcolor{#0f0}{b = -1}\textbf{, }\textcolor{red}{c = -1}$ $x = \frac{-(\textcolor{#0f0}{-1}) \pm \sqrt{(\textcolor{#0f0}{-1})^{2} - 4(\textcolor{#2192ff}{1})(\textcolor{red}{-1})}}{2(\textcolor{#2192ff}{1})}$ $x = \frac{1 \pm \sqrt{5}}{2}$ Earlier we said ‘let $y = \tan{x}$ ‘. We are going to use that equation since we want to solve for $x$ . When $\textcolor{#2192ff}{y}$ is $\textcolor{#2192ff}{\frac{1 + \sqrt{5}}{2}}$ ,
$\textcolor{#2192ff}{y} = \tan{x}$ $\textcolor{#2192ff}{\frac{1 + \sqrt{5}}{2}} = \tan{x}$ $x = \tan^{-1}\left(\frac{1 + \sqrt{5}}{2}\right)$ $x = 1.02 \textmd{ to 3 significant figures}$ When $\textcolor{#0f0}{y}$ is $\textcolor{#0f0}{\frac{1 - \sqrt{5}}{2}}$ ,
$\textcolor{#0f0}{y} = \tan{x}$ $\textcolor{#0f0}{\frac{1 - \sqrt{5}}{2}} = \tan{x}$ $x = \tan^{-1}\left(\frac{1 - \sqrt{5}}{2}\right)$ $x = -0.554 \textmd{ to 3 significant figures}$ Therefore, our final answer is,
$x = 1.02 \textbf{, }\ \ x = -0.554$

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