Quadratics

Intuition

Quadratics are the simplest curved relationships: Linear functions give straight lines — constant rate of change. Quadratics give parabolas — the rate of change itself changes. This makes them the natural next step after linear algebra. Every projectile follows a quadratic path, every area calculation involves quadratics, and many optimisation problems reduce to finding the vertex of a parabola.

Why it matters: Quadratics appear everywhere — the path of a ball, the shape of a satellite dish, the profit function of a business, the growth of a population. Understanding them gives you the tools to model any situation where something accelerates or decelerates.

The key insight: The discriminant ($b^2 - 4ac$) tells you everything about the roots without solving the equation. Positive means two real roots, zero means one repeated root, negative means no real roots. This is the quickest way to determine the nature of solutions.

Board Coverage

1. The Quadratic Function

Definition. A quadratic function is a function of the form:

$f(x) = ax^2 + bx + c$

Where $a, b, c \in \mathbb{R}$ and $a \neq 0$.

The graph of $y = ax^2 + bx + c$ is a parabola — a symmetrical U-shaped curve (opening upwards if $a > 0$Downwards if $a < 0$). The axis of symmetry is the vertical line $x = -\frac{b}{2a}$.

2. Completing the Square

The technique of completing the square rewrites a quadratic in the form $a(x - p)^2 + q$From which The vertex, axis of symmetry, and extremum are immediately readable.

Theorem. Every quadratic $ax^2 + bx + c$ with $a \neq 0$ can be written in the form $a(x - p)^2 + q$ for some $p, q \in \mathbb{R}$.

Proof. Factor out $a$ from the first two terms:

We seek to express $x^2 + \frac{b}{a}x$ as a perfect square plus a constant. Recall that:

$(x + d)^2 = x^2 + 2dx + d^2$

We need $2d = \frac{b}{a}$So $d = \frac{b}{2a}$. Then:

Substituting back:

Setting $p = -\frac{b}{2a}$ and $q = \frac{4ac - b^2}{4a}$We have $ax^2 + bx + c = a(x - p)^2 + q$. $\blacksquare$

Intuition (Geometric). Consider the expression $x^2 + bx$. This represents the area of a square of Side $x$ plus a rectangle of dimensions $x \times b$. We split the rectangle into two strips of $x \times \frac{b}{2}$And rearrange to form an L-shape. The “missing corner” to complete the Larger square is a small square of side $\frac{b}{2}$With area $\left(\frac{b}{2}\right)^2 = \frac{b^2}{4}$. We add and subtract this to preserve equality:

$x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}$

3. The Quadratic Formula

Theorem (Quadratic Formula). The solutions of $ax^2 + bx + c = 0$ (with $a \neq 0$) are:

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

Proof. We derive this by completing the square on $ax^2 + bx + c = 0$.

Intuition. The quadratic formula is nothing more than completing the square in fully symbolic Form. Every step is reversible, so the formula is necessary and sufficient: it gives all solutions And no extraneous ones.

4. The Discriminant

Definition. The discriminant of $ax^2 + bx + c = 0$ is:

$\Delta = b^2 - 4ac$

Theorem. The nature of the roots of $ax^2 + bx + c = 0$ is determined by the discriminant:

Proof. From the quadratic formula, the roots are $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

• If $\Delta > 0$: $\sqrt{\Delta}$ is a positive real number, giving two distinct real values.
• If $\Delta = 0$: both roots equal $\frac{-b}{2a}$A single repeated root.
• If $\Delta < 0$: $\sqrt{\Delta}$ is not a real number, so no real roots exist. (Complex roots exist but are beyond this course.) $\blacksquare$

4.1 Discriminant and Graph Shape

The discriminant directly determines the position of the parabola relative to the $x$-axis.

Theorem (Discriminant and Graph Position). Let $f(x) = ax^2 + bx + c$ with $a \neq 0$. In Completed-square form, the extremum value is $f\left(-\frac{b}{2a}\right) = -\frac{\Delta}{4a}$. Therefore:

• $\Delta > 0$: the extremum lies on the opposite side of the $x$-axis from the direction the parabola opens, so the graph crosses the $x$-axis at two distinct points.
• $\Delta = 0$: the vertex lies on the $x$-axis; the graph touches the axis at exactly one point.
• $\Delta < 0$: the vertex lies on the same side of the $x$-axis as the direction the parabola opens, so the graph never meets the axis.

Adjust sliders a, b, c to see how the quadratic $y = ax^2 + bx + c$ changes. Observe how the discriminant $\Delta = b^2 - 4ac$ determines whether the curve crosses, touches, or misses the $x$-axis.

Furthermore, when $\Delta < 0$:

Proof. From the completed-square form $f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a}$The minimum value (when $a > 0$) or Maximum value (when $a < 0$) equals $-\frac{\Delta}{4a}$Attained at $x = -\frac{b}{2a}$.

If $a > 0$ and $\Delta > 0$: the minimum is $-\frac{\Delta}{4a} < 0$So the vertex is below the $x$-axis. Since the parabola opens upward, it must cross the axis twice.

If $a > 0$ and $\Delta = 0$: the minimum is $0$So the vertex sits on the $x$-axis.

If $a > 0$ and $\Delta < 0$: the minimum is $-\frac{\Delta}{4a} > 0$So the vertex is above the $x$-axis, and $f(x) \geq -\frac{\Delta}{4a} > 0$ for all $x$.

The $a < 0$ cases follow by symmetry (or by applying the above to $-f(x)$). $\blacksquare$

4.2 Repeated Roots: Geometric Interpretation

Definition. When $\Delta = 0$The equation $ax^2 + bx + c = 0$ has a single repeated root $\alpha = -\frac{b}{2a}$. We say $\alpha$ is a double root (or root of multiplicity 2), and the Quadratic factors as $a(x - \alpha)^2 = 0$.

Theorem (Tangent at the Vertex). If $f(x) = ax^2 + bx + c$ has a repeated root at $\alpha$Then The tangent to $y = f(x)$ at $x = \alpha$ is the line $y = 0$ (the $x$-axis itself).

Proof. Since $\alpha$ is a repeated root, $f(\alpha) = 0$. The derivative is $f'(x) = 2ax + b$ And at the repeated root:

$f'(\alpha) = 2a\left(-\frac{b}{2a}\right) + b = -b + b = 0$

The tangent at $x = \alpha$ is $y - f(\alpha) = f'(\alpha)(x - \alpha)$Which gives $y = 0$. $\blacksquare$

This means that when $\Delta = 0$The $x$-axis is tangent to the parabola at the vertex. The Parabola “kisses” the axis at one point and bounces back, rather than crossing it.

5. Solving Quadratic Equations

5.1 By Factorisation

If $ax^2 + bx + c = 0$ can be written as $(px + q)(rx + s) = 0$Then by the zero product property, $x = -q/p$ or $x = -s/r$.

5.2 By Completing the Square

Useful when the quadratic doesn’t factorise and you want to understand the geometry.

5.3 By the Formula

Always works (when $\Delta \geq 0$), but can be computationally heavier.

6. Quadratic Inequalities

Solving $ax^2 + bx + c > 0$ (or < 0$$\geq 0$$\leq 0) requires understanding the sign of the Quadratic across the real line.

Method.

1. Find the roots of $ax^2 + bx + c = 0$.
2. Sketch the parabola (knowing whether $a > 0$ or $a < 0$).
3. Read off the regions where the quadratic is positive or negative.

Theorem. If $m < 0$ and $a < b$Then $ma > mb$.

Proof. From $a < b$We have $b - a > 0$. Since $m < 0$ and $b - a > 0$Their product $m(b - a) < 0$. So $mb - ma < 0$Giving $ma > mb$. $\blacksquare$

6.1 Rigorous Sign Chart Method

Theorem (Sign of a Factored Quadratic). Let $f(x) = a(x - \alpha)(x - \beta)$ with real roots $\alpha < \beta$ and $a \neq 0$. Then the sign of $f$ on each interval is determined by:

Proof. Consider $a > 0$. For $x > \beta$: both $(x - \alpha) > 0$ and $(x - \beta) > 0$So $f(x) > 0$. For $\alpha < x < \beta$: we have $(x - \alpha) > 0$ but $(x - \beta) < 0$So their Product is negative and $f(x) < 0$. For $x < \alpha$: both factors are negative, so their product is Positive and $f(x) > 0$. The $a < 0$ case reverses all signs. $\blacksquare$

Corollary. For $\Delta = 0$ (repeated root $\alpha$): $f(x) = a(x - \alpha)^2$ is always Non-negative when $a > 0$ and always non-positive when $a < 0$With equality only at $x = \alpha$.

Method (Systematic Sign Chart).

1. Rearrange the inequality to the form $f(x) \gtreqqless 0$.
2. Factorise $f(x)$ into linear factors if possible.
3. Identify all critical values: roots of the numerator, zeros of the denominator (if rational), and any points where $f$ is undefined.
4. Draw a sign chart: test one point from each interval between consecutive critical values.
5. Select the intervals that satisfy the original inequality, respecting strict ($>$, $<$) vs non-strict ($\geq$, $\leq$) conditions.

Tip: Tip never include points where the denominator is zero.

7. Simultaneous Equations (Linear-Quadratic)

When solving a system of one linear and one quadratic equation, we substitute the linear equation Into the quadratic.

Example. Solve simultaneously:

Setting them equal:

$x = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}$

The discriminant is positive, confirming two intersection points — which corresponds geometrically To the line cutting the parabola twice.

8. Quadratics in Disguise

Definition. A quadratic in disguise (or quadratic form equation) is an equation that is not Quadratic in its stated variable, but can be reduced to a quadratic equation by an appropriate Substitution.

Theorem. If an equation can be rewritten as $a\left[g(x)\right]^2 + b\left[g(x)\right] + c = 0$ For some expression $g(x)$ and constants $a \neq 0, b, c$Then the substitution $z = g(x)$ reduces It to the quadratic $az^2 + bz + c = 0$.

Common Patterns.

9. Vieta’s Formulas

Theorem (Vieta’s Formulas). If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ with $a \neq 0$Then:

$\alpha + \beta = -\frac{b}{a} \quad \mathrm{(Sum of Roots)}$

$\alpha\beta = \frac{c}{a} \quad \mathrm{(Product of Roots)}$

Proof. By the factor theorem, since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$We can Write:

$ax^2 + bx + c = a(x - \alpha)(x - \beta)$

Expanding the right-hand side:

$a(x - \alpha)(x - \beta) = a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] = ax^2 - a(\alpha + \beta)x + a\alpha\beta$

Equating coefficients with $ax^2 + bx + c$:

• Coefficient of $x$: $b = -a(\alpha + \beta)$Hence $\alpha + \beta = -\frac{b}{a}$.
• Constant term: $c = a\alpha\beta$Hence $\alpha\beta = \frac{c}{a}$.

$\blacksquare$

Corollary (Monic Case). For a monic quadratic $x^2 + bx + c = 0$:

$\alpha + \beta = -b \quad \mathrm{and} \quad \alpha\beta = c$

9.1 Relationship Between Roots and Coefficients

Vieta’s formulas allow us to deduce properties of the roots directly from the coefficients, without Solving the equation.

Theorem (Sign of Roots). Let $\alpha$ and $\beta$ be real roots of $ax^2 + bx + c = 0$ with $a > 0$ and $\Delta \geq 0$. Then:

Proof. We prove the first case; the others follow similarly.

Both roots positive means $\alpha + \beta > 0$ and $\alpha\beta > 0$. By Vieta: $\alpha + \beta = -\frac{b}{a}$So $-\frac{b}{a} > 0 \implies b < 0$ (since $a > 0$). Also $\alpha\beta = \frac{c}{a}$So $\frac{c}{a} > 0 \implies c > 0$.

Conversely, if $b < 0$ and $c > 0$Then $\alpha + \beta = -\frac{b}{a} > 0$ and $\alpha\beta = \frac{c}{a} > 0$. Since both the sum and product are positive, both roots must be Positive (if one were negative and the other positive, their product would be negative; if both were Negative, their sum would be negative). $\blacksquare$

9.2 Constructing Equations from Roots

Given two numbers $\alpha$ and $\beta$The monic quadratic with those roots is:

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$

More generally, the quadratic with leading coefficient $a$ is:

$ax^2 - a(\alpha + \beta)x + a\alpha\beta = 0$

9.3 Symmetric Functions of Roots

A symmetric function of $\alpha$ and $\beta$ is one that is unchanged when $\alpha$ and $\beta$ Are swapped. Vieta’s formulas allow us to evaluate many symmetric functions without finding the Roots individually.

Common Identities.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$

$\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$

10. Applying Quadratic Theory to Other Equations

Many equations that are not quadratic can be solved by recognising quadratic structure or By algebraic manipulation that produces a quadratic.

10.1 Exponential Equations

Method. For equations of the form $a \cdot p^{2x} + b \cdot p^x + c = 0$ where $p > 0$ Substitute $u = p^x$ (so $u > 0$) to obtain a quadratic in $u$. After solving, take logarithms to Recover $x$.

10.2 Trigonometric Equations

Method. Equations such as $a\cos^2\theta + b\cos\theta + c = 0$ are quadratic in $\cos\theta$. Solve for the trigonometric ratio, then find $\theta$ within the specified interval. Always check That the values fall within the valid range $[-1, 1]$.

10.3 Equations with Square Roots

Equations involving $\sqrt{f(x)}$ can sometimes be reduced to quadratics by isolating the radical And squaring both sides.

11. Problem Set

Problem 1. Write $3x^2 - 12x + 7$ in the form $a(x - p)^2 + q$And hence state the minimum Value and the value of $x$ at which it occurs.

If you get this wrong, revise: Completing the square

Problem 2. Solve $2x^2 + 5x - 4 = 0$Giving your answer in the form $a \pm b\sqrt{c}$.

If you get this wrong, revise: Quadratic formula

Problem 3. Find the range of values of $k$ for which $kx^2 - 6x + 4 = 0$ has real roots.

If you get this wrong, revise: Discriminant

Problem 4. Solve the inequality $x^2 - 3x - 10 > 0$.

If you get this wrong, revise: Quadratic inequalities

Problem 5. Solve the simultaneous equations $y = x - 1$ and $x^2 + y^2 = 13$.

If you get this wrong, revise: Simultaneous equations

Problem 6. The function $f(x) = px^2 + qx + r$ has a minimum value of $-5$ at $x = 2$And Passes through the point $(0, 7)$. Find $p$, $q$And $r$.

If you get this wrong, revise: Completing the square

Problem 7. Show that $x^2 + 4x + 9 > 0$ for all real $x$.

If you get this wrong, revise: Discriminant

Problem 8. Express $\frac{2x^2 - 8x + 5}{x - 3}$ in the form $Ax + B + \frac{C}{x - 3}$.

If you get this wrong, revise: Polynomial division

Problem 9. Find the equation of the tangent to the curve $y = x^2 - 4x + 3$ at the point where $x = 1$.

If you get this wrong, revise: Differentiation

Problem 10. Prove that the equation $x^2 + 2kx + k^2 + 1 = 0$ has no real roots for any real Value of $k$.

If you get this wrong, revise: Discriminant

Problem 11. A rectangle has length $(x + 5)$ cm and width $(x + 2)$ cm. The area is $30\mathrm{ cm}^2$. Find $x$.

If you get this wrong, revise: Quadratic formula

Problem 12. Given $f(x) = x^2 - 6x + 14$Find the range of $f$ and solve $f(x) \leq 10$.

If you get this wrong, revise: Completing the square and Quadratic inequalities

Problem 13. Solve $x^4 - 13x^2 + 36 = 0$.

If you get this wrong, revise: Quadratics in disguise

Problem 14. The roots of $3x^2 - 5x + 1 = 0$ are $\alpha$ and $\beta$. Without solving the Equation, find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.

If you get this wrong, revise: Vieta’s formulas

Problem 15. Solve $4^{2x} - 5 \cdot 4^x + 4 = 0$.