Quadratic Equations

1. Solution of Quadratic Equations

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a \neq 0$. Solving a quadratic equation means finding the values of $x$ that satisfy the equation. There are several methods for solving quadratic equations:

1.1 Factorisation

If the quadratic expression can be factorised into two linear factors, then the solutions can be found by setting each factor equal to zero. For example:

$x^2 + 5x + 6 = (x + 2)(x + 3) = 0$

Therefore, $x + 2 = 0$ or $x + 3 = 0$, which gives $x = -2$ or $x = -3$.

1.2 Completing the Square

Completing the square involves manipulating the quadratic equation to create a perfect square trinomial on one side. This allows us to take the square root of both sides and solve for $x$.

1. Divide the entire equation by $a$ (if $a \neq 1$).
2. Move the constant term to the right side of the equation.
3. Take half of the coefficient of the $x$ term, square it, and add it to both sides of the equation.
4. Factorise the left side as a perfect square.
5. Take the square root of both sides.
6. Solve for $x$.

1.3 Quadratic Formula

The quadratic formula provides a direct method for finding the solutions of any quadratic equation. The formula is:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The discriminant, $b^2 - 4ac$, determines the nature of the roots.

2. Nature of Roots

The discriminant, $b^2 - 4ac$, determines the nature of the roots of a quadratic equation:

Discriminant	Nature of Roots
$b^2 - 4ac > 0$	Two distinct real roots
$b^2 - 4ac = 0$	One real root (repeated)
$b^2 - 4ac < 0$	Two complex conjugate roots

3. Quadratic Inequalities

Quadratic inequalities are inequalities involving quadratic expressions. To solve a quadratic inequality, we first find the roots of the corresponding quadratic equation ($ax^2 + bx + c = 0$). These roots divide the number line into intervals. We then test a value from each interval to determine whether the quadratic expression is positive or negative in that interval.

For example, consider the inequality $x^2 - 5x + 6 > 0$. The roots of the equation $x^2 - 5x + 6 = 0$ are $x = 2$ and $x = 3$. We test the intervals $(-\infty, 2)$, $(2, 3)$, and $(3, \infty)$.

• Interval $(-\infty, 2)$: Let $x = 0$. Then $0^2 - 5(0) + 6 = 6 > 0$. So, the inequality holds for this interval.
• Interval $(2, 3)$: Let $x = 2.5$. Then $(2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 < 0$. So, the inequality does not hold for this interval.
• Interval $(3, \infty)$: Let $x = 4$. Then $4^2 - 5(4) + 6 = 16 - 20 + 6 = 2 > 0$. So, the inequality holds for this interval.

Therefore, the solution to the inequality $x^2 - 5x + 6 > 0$ is $x < 2$ or $x > 3$. In interval notation, this is $(-\infty, 2) \cup (3, \infty)$.