Introduction and Quadratic Equations

What is a Quadratic Equation?

In the last chapter, polynomials of degree 2 were called quadratic. When we set a quadratic polynomial equal to zero, we get a quadratic equation.

A quadratic equation in the variable $x$ is any equation that can be written in the standard form:

$$ax^2 + bx + c = 0, \quad a \neq 0$$

where $a$, $b$, $c$ are real numbers and — crucially — $a \neq 0$. (If $a = 0$, the $x^2$ term vanishes and it is no longer quadratic.)

Key Point: The defining feature is the $x^2$ term with a non-zero coefficient. The highest power of $x$ must be exactly 2.

Examples: $x^2 - 5x + 6 = 0$, $2x^2 = 8$, $3x^2 + x = 0$ are all quadratic equations.

[Board Important] Always rearrange an equation into standard form $ax^2 + bx + c = 0$ before identifying $a$, $b$, $c$ or applying any method.

Checking Whether an Equation is Quadratic

Some equations look complicated but turn out to be quadratic (or not) once simplified. The test: bring everything to one side, simplify, and check if the highest power of $x$ is exactly 2.

Worked outline

• $(x + 1)^2 = 2(x - 3)$: expand → $x^2 + 2x + 1 = 2x - 6$ → $x^2 + 7 = 0$. Highest power 2 → quadratic.
• $x(x + 1) + 8 = (x + 2)(x - 2)$: → $x^2 + x + 8 = x^2 - 4$ → $x + 12 = 0$. The $x^2$ cancels → not quadratic (it's linear).

Key Point: After full simplification, the equation is quadratic only if the $x^2$ term survives with a non-zero coefficient. Don't judge from the unsimplified form.

[Board Important] A favourite trap: an equation with $x^2$ on both sides may simplify to a linear equation. Always simplify first.

Roots of a Quadratic Equation

A root (or solution) of $ax^2 + bx + c = 0$ is a value of $x$ that makes the equation true, i.e. $a\alpha^2 + b\alpha + c = 0$ for $x = \alpha$.

• A quadratic equation has at most two roots (since its degree is 2).
• These roots are exactly the zeroes of the polynomial $ax^2 + bx + c$ — the same idea from Chapter 2, now applied to solving.

Checking a root

Is $x = 2$ a root of $x^2 - 3x + 2 = 0$? Substitute: $(2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0$. Yes, 2 is a root.

Key Point: 'Solve the quadratic equation' means 'find its roots' — the values of $x$ satisfying it.

[Board Important] To verify a given value is a root, substitute it and check the equation reduces to 0. This is a common 1-mark task.

Forming Quadratic Equations from Situations

Many problems describe a real-life situation that translates into a quadratic equation. The skill is to assign a variable and express the condition.

Worked outline

'The product of two consecutive positive integers is 156.'

1. Let the smaller integer be $x$; the next is $x + 1$.
2. Their product: $x(x + 1) = 156$.
3. Expand: $x^2 + x - 156 = 0$ — a quadratic equation.

Area example

'The length of a rectangular plot is 3 m more than its breadth, and its area is 70 m².'

• Let breadth $= x$. Length $= x + 3$. Area: $x(x + 3) = 70 \Rightarrow x^2 + 3x - 70 = 0$.

Key Point: Define the variable, write the condition as a product or expression, then bring it to standard form.

[Board Important] State 'Let … be $x$' clearly with units. Forming the correct equation is often worth a mark or two even before solving.

Solved Examples

Example 1: Is it a quadratic equation?

Is $(x - 2)^2 + 1 = 2x - 3$ a quadratic equation?

Solution:

1. Expand: $x^2 - 4x + 4 + 1 = 2x - 3$.
2. Bring to one side: $x^2 - 4x + 5 - 2x + 3 = 0 \Rightarrow x^2 - 6x + 8 = 0$.
3. Highest power is 2.

Final Answer: Yes, it is quadratic.

Takeaway: Simplify fully, then check the $x^2$ term survives.

Example 2: Not a quadratic

Is $x(x + 1) + 8 = (x + 2)(x - 2)$ a quadratic equation?

Solution:

1. Expand: $x^2 + x + 8 = x^2 - 4$.
2. The $x^2$ terms cancel: $x + 8 = -4 \Rightarrow x + 12 = 0$.
3. Highest power is 1.

Final Answer: No, it is linear, not quadratic.

Takeaway: Equal $x^2$ on both sides can cancel — always simplify.

Example 3: Identify $a$, $b$, $c$

Write $3x^2 = 5x - 2$ in standard form and state $a$, $b$, $c$.

Solution:

1. Move all terms to one side: $3x^2 - 5x + 2 = 0$.
2. So $a = 3$, $b = -5$, $c = 2$.

Final Answer: $a = 3$, $b = -5$, $c = 2$.

Takeaway: Rearrange to $ax^2 + bx + c = 0$ before reading coefficients.

Example 4: Check a root

Is $x = -3$ a root of $x^2 + x - 6 = 0$?

Solution:

1. Substitute: $(-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0$.
2. Since it equals 0, $-3$ is a root.

Final Answer: Yes, $x = -3$ is a root.

Takeaway: Substitute and check the equation becomes 0.

Example 5: Form an equation (consecutive integers)

The product of two consecutive positive integers is 156. Form the quadratic equation.

Solution:

1. Let the integers be $x$ and $x + 1$.
2. $x(x + 1) = 156 \Rightarrow x^2 + x - 156 = 0$.

Final Answer: $x^2 + x - 156 = 0$.

Takeaway: Consecutive integers are $x$ and $x + 1$.

Example 6: Form an equation (area)

The length of a rectangular hall is 3 m more than twice its breadth. Its area is 90 m². Form the equation.

Solution:

1. Let breadth $= x$. Length $= 2x + 3$.
2. Area: $x(2x + 3) = 90 \Rightarrow 2x^2 + 3x - 90 = 0$.

Final Answer: $2x^2 + 3x - 90 = 0$.

Takeaway: Translate 'more than twice' as $2x + 3$, then use area = length × breadth.

Example 7: Find $k$ given a root

If $x = 2$ is a root of $x^2 - kx + 2 = 0$, find $k$.

Solution:

1. Substitute $x = 2$: $(2)^2 - k(2) + 2 = 0 \Rightarrow 4 - 2k + 2 = 0$.
2. $6 = 2k \Rightarrow k = 3$.

Final Answer: $k = 3$.

Takeaway: A known root, substituted, gives an equation for the unknown coefficient.

Example 8: Form an equation (ages)

A boy's age two years ago, multiplied by his age in three years, is 90. Form the equation (let present age $= x$).

Solution:

1. Two years ago: $x - 2$. In three years: $x + 3$.
2. Product: $(x - 2)(x + 3) = 90 \Rightarrow x^2 + x - 6 = 90 \Rightarrow x^2 + x - 96 = 0$.

Final Answer: $x^2 + x - 96 = 0$.

Takeaway: Express each age relative to the present, then multiply.

Example 9: Standard form with fractions

Write $\dfrac{x^2}{2} - \dfrac{x}{3} = 1$ in standard form.

Solution:

1. Multiply throughout by 6: $3x^2 - 2x = 6$.
2. Bring to one side: $3x^2 - 2x - 6 = 0$.

Final Answer: $3x^2 - 2x - 6 = 0$.

Takeaway: Clear fractions by multiplying by the LCM of denominators first.

Example 10: Decide quadratic with a parameter

For what value of $a$ is $(a - 1)x^2 + 3x + 2 = 0$ NOT a quadratic equation?

Solution:

1. It fails to be quadratic when the coefficient of $x^2$ is 0.
2. $a - 1 = 0 \Rightarrow a = 1$.

Final Answer: $a = 1$.

Takeaway: Quadratic requires the $x^2$ coefficient $\neq 0$; set it to 0 to find the excluded value.