Understanding equations that can be reduced to quadratic form is essential. These equations, often appearing complex at first glance, can be transformed into a quadratic equation through substitution. This topic is particularly relevant for problems involving higher powers and trigonometric functions.

Recognising Equations Reducible to Quadratics

Identifying an equation that can be transformed into a quadratic form involves looking for patterns akin to the standard quadratic equation $ax^2 + bx + c = 0$. These patterns may involve variables raised to different powers or variables within trigonometric functions.

Key Indicators:

• Terms that can be squared or involve square roots.
• Equations with trigonometric functions where the variable appears in different forms, like $\sin(x)$ and $\sin^2(x)$.
• Higher power terms where one term is a power of another, such as $x^4$and $x^2$.

Methods of Substitution

Substitution simplifies these equations into a quadratic form. The process involves replacing a complex part of the equation with a simpler variable, typically $u$, to transform it into a quadratic equation.

Steps for Substitution:

1. Identify the Substitution: Choose a part of the equation that, when substituted, will simplify it to a quadratic form.

2. Rewrite the Equation: Replace the identified part with $u$ and rewrite the equation.

3. Solve the Quadratic Equation: Solve the new equation as a standard quadratic.

4. Back-Substitute: Replace $u$ with its original expression and solve for $x$.

Examples:

Example 1: Higher Power Equation

Solve $x^4 - 5x^2 + 6 = 0$

Solution:

1. Substitute: Let $u = x^2$. The equation becomes $u^2 - 5u + 6 = 0$.

2. Factor: $(u - 2)(u - 3) = 0$, so $u = 2$ or $u = 3$.

3. Reverse: Put $x^2$ back for $u$, getting $x^2 = 2$ and $x^2 = 3$.

4. Solve: For $x^2 = 2$, $x = \pm\sqrt{2}$. For $x^2 = 3$, $x = \pm\sqrt{3}$.

Answers: $x = \pm\sqrt{2}$, $x = \pm\sqrt{3}$.

Example 2: Trigonometric Function

Solve $\sin^2(x) - 3\sin(x) + 2 = 0.$

Solution:

1. Substitute: Let $u = \sin(x)$. Equation is $u^2 - 3u + 2 = 0$.

2. Factor: $(u - 1)(u - 2) = 0$, thus $u = 1$ or $u = 2$.

3. Reject: $u = 2$ is invalid as $\sin(x)$ is between -1 and 1.

4. Reverse: For $u = 1$, $\sin(x) = 1$.

5. Solve: $x = \frac{\pi}{2} + 2k\pi$, where $k$ is an integer.

Answer: $x = \frac{\pi}{2} + 2k\pi$

Advanced Techniques

Working with Higher Powers

In equations with higher powers, such as $x^6 - 2x^3 + 1 = 0$, the substitution method still applies. Here, setting $u = x^3$ transforms the equation into a quadratic form. Solving the quadratic equation and then back-substituting $u$with $x^3$ leads to the original variable solutions.

Trigonometric Transformations

For equations involving trigonometric functions, like $\cos^2(x) - 4\cos(x) + 3 = 0$, the substitution $u = \cos(x)$ can be used.

After solving the quadratic in $u$, the solutions in terms of $x$ can be found using trigonometric identities and inverse functions.

Examples:

Example 1: Trigonometric Equation

Solve $4\cos^2(x) - 3\cos(x) - 1 = 0$ for $x$.

Solution:

1. Let $u = \cos(x)$, so $4u^2 - 3u - 1 = 0$.

2. Apply quadratic formula: $u = \frac{3 \pm \sqrt{25}}{8}$.

3. Simplify to find $u = 1$ or $u = -\frac{1}{4}$.

4. For $u = 1$, $x = 2\pi n$ (since $\cos(x) = 1$ at these points).

5. For $u = -\frac{1}{4}$, $x = \arccos(-\frac{1}{4}) + 2\pi n$.

Answers: $x = 2\pi n$ and $x = \arccos(-\frac{1}{4}) + 2\pi n$, for any integer $n$.

Example 2: Higher Power Equation

Solve $x^6 - 5x^3 + 6 = 0$ for $x$.

Solution:

1. Let $u = x^3$, changing the equation to $u^2 - 5u + 6 = 0$.

2. Factorise to $(u - 2)(u - 3) = 0$, giving $u = 2$ or $u = 3$.

3. Replace $u$ with $x^3$ to get $x^3 = 2$ and $x^3 = 3$.

4. Solve each to find $x = \sqrt[3]{2}$ and $x = \sqrt[3]{3}$.

Answers: $x = \sqrt[3]{2}$ and $x = \sqrt[3]{3}$.