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

Private: CAIE Alevel Pure Maths 1 (AS)

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

e.g. $x^4 - 5x^2 + 4 = 0$ , $6x +\sqrt{x} - 1 = 0$ , $\tan^2 x = 1 + \tan x$ .

Equations that are quadratic in some function of $x$ have the form $ax^2 + bx + c = 0$ , where the function of $x$ is raised to the second power.

To solve these equations, we use a technique called substitution, where we replace the function of $x$ with a new variable, and solve for that variable.

Example 1

Solve the equation $x^4 - 5x^2 + 4 = 0$ .

We notice that this equation is quadratic in $x^2$ .
Therefore, we can substitute $y = x^2$ , and rewrite the equation as $y^2 - 5y + 4 = 0$ . This equation is now a quadratic equation that can be solved by factoring or by using the quadratic formula.
We get: $y^2 - 5y + 4 = (y - 4)(y - 1) = 0$
Therefore, $y = 4$ or $y = 1$ .
Substituting back $y = x^2$ , we get: $x^2 = 4$ or $x^2 = 1$
Solving for $x$ , we get: $x = \pm2$ or $x = \pm1$

Example 2

Solve the equation $6x + \sqrt{x} - 1 = 0$ .

We notice that this equation is quadratic in $\sqrt{x}$ .
Therefore, we can substitute $y = \sqrt{x}$ , and rewrite the equation as $6y^2 + y - 1 = 0$ . This equation is now a quadratic equation that can be solved by factoring or by using the quadratic formula.
We get: $(2y + 1)(3y - 1) = 0$
Therefore, $y = -\frac{1}{2}$ or $y = \frac{1}{3}$ .
Substituting back $y = \sqrt{x}$ , we get: $\sqrt{x} = -\frac{1}{2}$ or $\sqrt{x} = \frac{1}{3}$
Since $\sqrt{x}$ can’t be negative, we ignore the first solution. Solving for $x$ , we get: $x = (1/3)^2 = 1/9$ .

Example 3

Solve the equation $\tan^2x = 1 + \tan x$ , $x \in (0, \pi)$

We notice that this equation is quadratic in $\tan x$ .
Therefore, we can substitute $y = \tan x$ , and rewrite the equation as $y^2 - y - 1 = 0$ . This equation is now a quadratic equation that can be solved by using the quadratic formula.
We get: $y = \frac{1 ± \sqrt{5}}{2}$
Substituting back $y = \tan x$ , we get: $\tan x = \frac{1 ± \sqrt{5}}{2}$
Using the inverse tangent function, we get:
$x = \arctan(\frac{1 + \sqrt{5}}{2})$ or $x = \arctan\frac{1 - \sqrt{5}}{2} + π$ .
Therefore, we have two solutions for $x$ .

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