In this section, we'll explore various techniques for solving quadratic equations. These include factorisation, completing the square, and the quadratic formula. Additionally, we'll examine how to handle equations in quadratic form, which, although not inherently quadratic, can be transformed into quadratic equations.

**Solving Equations in Quadratic Form**

Some equations, while not quadratic initially, can be transformed into quadratic equations. This is achieved by substituting a function of $x$ into the equation.

**Example 1:**

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

**Solution:**

Let $u = x^2$, then the equation becomes $u^2 - 5u + 4 = 0$. Solving for $u$gives:

$(u - 4)(u - 1) = 0 \Rightarrow u = 4, 1$Thus, $x = \pm \sqrt{u} \Rightarrow x = \pm 2, \pm 1$.

**Example 2:**

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

**Solution:**

Let $u = \sqrt{x}$, transforming the equation to $6u^2 + u - 1 = 0$. Solving for $u$ gives:

$(3u - 1)(2u + 1) = 0 \Rightarrow u = \frac{1}{3}, -\frac{1}{2}$Since $x = u^2$, we reject $u = -\frac{1}{2}$ as it yields no real solutions. Therefore, $x = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.

**Factorising Quadratic Equations**

Factorising is effective when the quadratic can be easily broken down into factors.

**Example 1:**

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

**Solution:**

**Example 2:**

Solve $x^2 - 4x - 5 = 0$.

**Solution: **

**Completing the Square**

This method is useful for quadratic equations that are not easily factorisable.

**Example 1:**

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

**Solution:**

**Example 2:**

Solve $x^2 - 6x + 8 = 0$.

**Solution:**

**Using the Quadratic Formula**

This formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, can solve any quadratic equation.

**Example 1:**

Solve $2x^2 - 3x + 1 = 0$.

**Solution:**

**Example 2:**

Solve $x^2 - x - 2 = 0$.

**Solution:**

$x = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \Rightarrow x = 2, -1$