Quadratic Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Quadratic Equations — Quick Facts

Standard Form: $$ax^2 + bx + c = 0 \text{ where } a \neq 0$$

Quadratic Formula (for solving): $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Discriminant: $D = b^2 - 4ac$

Value of $D$	Nature of Roots
$D > 0$ (perfect square)	Two distinct rational roots
$D > 0$ (not perfect square)	Two distinct irrational roots
$D = 0$	Two equal real roots ($x = -\frac{b}{2a}$)
$D < 0$	No real roots (complex conjugates)

Sum and Product of Roots:

• Sum: $\alpha + \beta = -\frac{b}{a}$
• Product: $\alpha\beta = \frac{c}{a}$

Factorisation Method: If $ax^2 + bx + c = (px + q)(rx + s) = 0$, then $x = -\frac{q}{p}$ or $x = -\frac{s}{r}$

⚡ JAMB Exam Tip: Always check the discriminant first to determine root nature. When $D = 0$, the repeated root is $x = -\frac{b}{2a}$. Don’t use the quadratic formula when factorisation is quicker.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Quadratic Equations — Study Guide

Factorisation Worked Example:

Solve $2x^2 + 5x - 3 = 0$

Find two numbers that multiply to $2 \times (-3) = -6$ and add to $+5$. These are $+6$ and $-1$.

Rewrite: $2x^2 + 6x - x - 3 = 0$ Factor by grouping: $2x(x + 3) - 1(x + 3) = 0$ $(2x - 1)(x + 3) = 0$

$x = \frac{1}{2}$ or $x = -3$

Quadratic Formula Worked Example:

Solve $x^2 - 4x + 1 = 0$

$a = 1$, $b = -4$, $c = 1$ $D = (-4)^2 - 4(1)(1) = 16 - 4 = 12 > 0$

$x = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$

Forming Equations from Roots:

If roots are $\alpha$ and $\beta$: $$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Example: Form equation with roots 3 and -2.

Sum = $1$, Product = $-6$ $x^2 - x - 6 = 0$

Nature of Roots Conditions:

• Real roots: $b^2 - 4ac \geq 0$
• Equal roots: $b^2 = 4ac$
• Opposite signs (product negative): $ac < 0$
• Same sign: $ac > 0$
• Reciprocal roots: If roots are $\alpha, \beta$, then equation is $c x^2 + bx + a = 0$
• Equal magnitude but opposite sign: $b = 0$, so $ax^2 + c = 0$

⚡ Common Student Mistake: When forming an equation with given roots, don’t forget to expand completely. A common error is writing $x^2 - (\alpha + \beta) = 0$ (missing the product term).

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Quadratic Equations — Comprehensive Notes

Derivation of Quadratic Formula:

From $ax^2 + bx + c = 0$:

1. Divide by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
2. Complete the square: $x^2 + \frac{b}{a}x = -\frac{c}{a}$
3. Add $(\frac{b}{2a})^2$ to both sides: $(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$
4. Take square root: $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$
5. Solve: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Maximum and Minimum Values:

For $y = ax^2 + bx + c$:

• If $a > 0$: minimum at $x = -\frac{b}{2a}$, $y_{\min} = \frac{4ac - b^2}{4a}$
• If $a < 0$: maximum at $x = -\frac{b}{2a}$, $y_{\max} = \frac{4ac - b^2}{4a}$

Example: Find maximum value of $f(x) = -2x^2 + 8x - 3$

$x = -\frac{8}{2(-2)} = 2$ $f(2) = -2(4) + 8(2) - 3 = -8 + 16 - 3 = 5$

Condition for Common Root:

Two quadratics $ax^2 + bx + c = 0$ and $a’x^2 + b’x + c’ = 0$ share:

• One root in common if $\frac{a}{a’} = \frac{b}{b’} = \frac{c}{c’}$
• Or use: $(ac’ - a’c)(b’c - bc’) = (b c’ - b’c)(a’c - ac’)$

Range of $x$ for Inequalities:

For $x^2 - 5x + 6 < 0$: Factor: $(x-2)(x-3) < 0$

Sign analysis: $x < 2$ or $x > 3$ gives positive; $2 < x < 3$ gives negative.

Therefore solution: $2 < x < 3$

Simultaneous Equations with Quadratics:

Example: Solve $x + y = 5$ and $xy = 6$

From first: $y = 5 - x$ Substitute: $x(5 - x) = 6$ $5x - x^2 = 6$ $x^2 - 5x + 6 = 0$ $(x-2)(x-3) = 0$

$x = 2, y = 3$ or $x = 3, y = 2$

Word Problem Setup:

Example: A rectangular garden is 4m longer than its width. Its area is 96 m². Find dimensions.

Let width = $w$, length = $w + 4$ $w(w + 4) = 96$ $w^2 + 4w - 96 = 0$ $(w + 12)(w - 8) = 0$ $w = 8$ m (width), length = $12$ m

JAMB Pattern Analysis (2015-2024):

• 2015: Discriminant and nature of roots
• 2017: Forming equation from given roots
• 2019: Solving by factorisation
• 2021: Maximum/minimum value problems
• 2023: Solving simultaneous equations (linear + quadratic)
• 2024: Quadratic inequalities

⚡ Exam Strategy: When asked “find the range of values of $x$” for a quadratic inequality, first solve the equality, then use sign testing or sketch the parabola. Remember: if $a > 0$, the parabola opens upward, so the expression is negative between the roots.