Quadratics

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Completing the square, discriminant, inequalities
Edexcel	P1	Same core content
OCR (A)	Paper 1	Includes set notation for solution sets
CIE (9709)	P1	Quadratic equations, discriminant, inequalities

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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 $x = -\frac{b}{2a}$.

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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:

$$\begin{aligned} Ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x\right) + c \end{aligned}$$

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:

$$\begin{aligned} X^2 + \frac{b}{a}x &= \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} \end{aligned}$$

Substituting back:

$$\begin{aligned} Ax^2 + bx + c &= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\ &= a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a} \end{aligned}$$

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:

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

Example

Write $2x^2 - 12x + 22$ in completed square form.

$$\begin{aligned} 2x^2 - 12x + 22 &= 2(x^2 - 6x) + 22 \\ &= 2\left[(x - 3)^2 - 9\right] + 22 \\ &= 2(x - 3)^2 - 18 + 22 \\ &= 2(x - 3)^2 + 4 \end{aligned}$$

The vertex is $(3, 4)$And since the coefficient of $(x-3)^2$ is positive, the minimum value is $4$.

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3. The Quadratic Formula

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

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

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

$$\begin{aligned} Ax^2 + bx + c &= 0 \\ X^2 + \frac{b}{a}x &= -\frac{c}{a} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} &= -\frac{c}{a} \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{c}{a} \\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \\ X + \frac{b}{2a} &= \pm\frac◆LB◆\sqrt{b^2 - 4ac}◆RB◆◆LB◆2a◆RB◆ \\ X &= \frac◆LB◆-b \pm \sqrt{b^2 - 4ac}◆RB◆◆LB◆2a◆RB◆ \quad \blacksquare \end{aligned}$$

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.

:::info Edexcel provides the quadratic formula in the formula booklet. AQA and OCR (A) do not — you Must memorise it. :::

:::tip Before applying the formula, check whether the equation can be solved more by Factorisation. Always check the discriminant first. :::

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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:

Condition	Number of Roots	Nature of Roots
$\Delta > 0$	2	Two distinct real roots
$\Delta = 0$	1	One repeated real root
$\Delta < 0$	0	No real roots (two complex conjugate roots)

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

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

Example

Find the set of values of $k$ for which $x^2 + 4x + k = 0$ has two distinct real roots.

$$\begin{aligned} \Delta &= 16 - 4k > 0 \\ 16 - 4k &> 0 \\ K &< 4 \end{aligned}$$

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 $f\left(-\frac{b}{2a}\right) = -\frac◆LB◆\Delta◆RB◆◆LB◆4a◆RB◆$. 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 $\Delta < 0$:

Condition	Conclusion
$a > 0$ and $\Delta < 0$	$f(x) > 0$ for all $x \in \mathbb{R}$ (parabola entirely above)
$a < 0$ and $\Delta < 0$	$f(x) < 0$ for all $x \in \mathbb{R}$ (parabola entirely below)

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

If $a > 0$ and $\Delta > 0$: the minimum is $-\frac◆LB◆\Delta◆RB◆◆LB◆4a◆RB◆ < 0$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 $-\frac◆LB◆\Delta◆RB◆◆LB◆4a◆RB◆ > 0$So the vertex is above the $x$-axis, and $f(x) \geq -\frac◆LB◆\Delta◆RB◆◆LB◆4a◆RB◆ > 0$ 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.

Example

Show that $4x^2 - 12x + 9 = 0$ has a repeated root, and verify that the $x$-axis is tangent at that point.

$\Delta = (-12)^2 - 4(4)(9) = 144 - 144 = 0$Confirming a repeated root.

The repeated root is $x = \frac{12}{8} = \frac{3}{2}$.

Verification: $f\left(\frac{3}{2}\right) = 4\left(\frac{9}{4}\right) - 12\left(\frac{3}{2}\right) + 9 = 9 - 18 + 9 = 0$.

$f'(x) = 8x - 12$So $f'\left(\frac{3}{2}\right) = 12 - 12 = 0$. The tangent at $x = \frac{3}{2}$ Is $y = 0$Confirming the $x$-axis is tangent to the curve.

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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.

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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.

:::caution A critical error in inequalities: when multiplying or dividing both sides by a negative Number, you must reverse the inequality sign. This is because multiplication by $-1$ is Order-reversing: if $a < b$Then $-a > -b$. :::

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 $ma > mb$. $\blacksquare$

Example

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

Factorise: $(x - 2)(x - 3) < 0$.

The parabola opens upwards (coefficient of $x^2$ is positive). It is negative between the roots:

$2 < x < 3$

Example

Solve $\frac{x + 1}{x - 2} \leq 3$.

$$\begin{aligned} \frac{x + 1}{x - 2} - 3 &\leq 0 \\ \frac{x + 1 - 3(x - 2)}{x - 2} &\leq 0 \\ \frac{x + 1 - 3x + 6}{x - 2} &\leq 0 \\ \frac{-2x + 7}{x - 2} &\leq 0 \\ \frac{2x - 7}{x - 2} &\geq 0 \end{aligned}$$

Critical values: $x = \frac{7}{2}$ (numerator zero) and $x = 2$ (denominator zero, undefined).

Sign analysis:

Interval	$2x - 7$	$x - 2$	Quotient
$x < 2$	$-$	$-$	$+$
$2 < x < 7/2$	$-$	$+$	$-$
$x > 7/2$	$+$	$+$	$+$

The quotient is $\geq 0$ when $x \leq \frac{7}{2}$ (including equality) but $x \neq 2$.

Solution: $x \leq \frac{7}{2}$, $x \neq 2$I.e., $x \in (-\infty, 2) \cup \left[\frac{7}{2}, \infty\right)$.

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:

Interval	Sign of $f$ when $a > 0$	Sign of $f$ when $a < 0$
$x < \alpha$	$+$	$-$
$\alpha < x < \beta$	$-$	$+$
$x > \beta$	$+$	$-$

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. :::

Example

Solve $-2x^2 + 3x + 5 \geq 0$ using a sign chart.

Factorise: $-2x^2 + 3x + 5 = -(2x^2 - 3x - 5) = -(2x - 5)(x + 1)$.

Critical values: $x = \frac{5}{2}$ and $x = -1$.

Sign chart for $g(x) = (2x - 5)(x + 1)$ (then apply the leading minus sign):

Interval	$2x - 5$	$x + 1$	$g(x)$	$-g(x)$
$x < -1$	$-$	$-$	$+$	$-$
$-1 < x < 5/2$	$-$	$+$	$-$	$+$
$x > 5/2$	$+$	$+$	$+$	$-$

We need $-g(x) \geq 0$Which occurs for $-1 \leq x \leq \frac{5}{2}$.

Solution: $x \in \left[-1, \frac{5}{2}\right]$.

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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:

$$\begin{aligned} Y &= 2x + 1 \\ Y &= x^2 + x - 3 \end{aligned}$$

Setting them equal:

$$\begin{aligned} 2x + 1 &= x^2 + x - 3 \\ X^2 - x - 4 &= 0 \end{aligned}$$

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

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

:::tip The discriminant of the resulting quadratic tells you the number of intersection points Between a line and a parabola: $\Delta > 0$ means 2 intersections, $\Delta = 0$ means tangent, $\Delta < 0$ means no intersection. :::

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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.

Original Form	Substitution	Reduced Equation
$ax^4 + bx^2 + c = 0$	$z = x^2$	$az^2 + bz + c = 0$
$a \cdot p^{2x} + b \cdot p^x + c = 0$	$z = p^x$	$az^2 + bz + c = 0$
$a\cos^2\theta + b\cos\theta + c = 0$	$z = \cos\theta$	$az^2 + bz + c = 0$
$a\sin^2\theta + b\sin\theta + c = 0$	$z = \sin\theta$	$az^2 + bz + c = 0$
$a\left(x + \frac{1}{x}\right)^2 + b\left(x + \frac{1}{x}\right) + c = 0$	$z = x + \frac{1}{x}$	$az^2 + bz + c = 0$

:::caution After solving the reduced quadratic for $z$You must substitute back to find $x$. Discard any values of $z$ that are incompatible with the substitution (e.g., $z = x^2$ requires $z \geq 0$, $z = p^x$ requires $z > 0$). Always verify solutions in the original equation. :::

Example

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

Let $z = x^2$ (note $z \geq 0$). Then $z^2 - 13z + 36 = 0$.

Factorising: $(z - 4)(z - 9) = 0$So $z = 4$ or $z = 9$.

Substituting back: $x^2 = 4 \implies x = \pm 2$And $x^2 = 9 \implies x = \pm 3$.

Solution: $x \in \{-3, -2, 2, 3\}$.

Example

Solve $x + 2 - \frac{3}{x} = 0$.

Multiply through by $x$ (noting $x \neq 0$):

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

$(x + 3)(x - 1) = 0 \implies x = -3 \mathrm{ or } x = 1$

Now consider the related equation $x^2 + \frac{4}{x^2} + 2\left(x + \frac{1}{x}\right) = 0$.

Let $z = x + \frac{1}{x}$. Then $z^2 = x^2 + 2 + \frac{1}{x^2}$So $x^2 + \frac{1}{x^2} = z^2 - 2$.

The equation becomes: $z^2 - 2 + 2z = 0$I.e., $z^2 + 2z - 2 = 0$.

$z = \frac◆LB◆-2 \pm \sqrt{4 + 8}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-2 \pm 2\sqrt{3}◆RB◆◆LB◆2◆RB◆ = -1 \pm \sqrt{3}$

For each value of $z$Solve $x + \frac{1}{x} = z$I.e., $x^2 - zx + 1 = 0$ by the quadratic Formula. This yields four solutions in total (two for each value of $z$), provided each resulting Discriminant is non-negative.

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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:

Condition on Roots	Condition on Coefficients	Via Vieta
Both roots positive	$b < 0$ and $c > 0$	$\alpha + \beta > 0$ and $\alpha\beta > 0$
Both roots negative	$b > 0$ and $c > 0$	$\alpha + \beta < 0$ and $\alpha\beta > 0$
Roots of opposite sign	$c < 0$	$\alpha\beta < 0$
One root is zero	$c = 0$	$\alpha\beta = 0$

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:

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

Example

Find the quadratic equation with roots $\alpha = 3$ and $\beta = -5$.

Sum: $\alpha + \beta = -2$. Product: $\alpha\beta = -15$.

Monic equation: $x^2 - (-2)x + (-15) = 0$I.e., $x^2 + 2x - 15 = 0$.

Verification: $(x - 3)(x + 5) = x^2 + 2x - 15$. Correct.

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.

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

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

$\frac◆LB◆1◆RB◆◆LB◆\alpha◆RB◆ + \frac◆LB◆1◆RB◆◆LB◆\beta◆RB◆ = \frac◆LB◆\alpha + \beta◆RB◆◆LB◆\alpha\beta◆RB◆$

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

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

Example

The roots of $2x^2 - 8x + 3 = 0$ are $\alpha$ and $\beta$. Without solving the equation, find $\alpha^2 + \beta^2$ and $\frac◆LB◆1◆RB◆◆LB◆\alpha◆RB◆ + \frac◆LB◆1◆RB◆◆LB◆\beta◆RB◆$.

From Vieta: $\alpha + \beta = \frac{8}{2} = 4$ and $\alpha\beta = \frac{3}{2}$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 16 - 3 = 13$.

$\frac◆LB◆1◆RB◆◆LB◆\alpha◆RB◆ + \frac◆LB◆1◆RB◆◆LB◆\beta◆RB◆ = \frac◆LB◆\alpha + \beta◆RB◆◆LB◆\alpha\beta◆RB◆ = \frac{4}{3/2} = \frac{8}{3}$.

Example

The roots of $x^2 - 5x + 2 = 0$ are $\alpha$ and $\beta$. Find a quadratic equation whose roots are $\alpha^2$ and $\beta^2$.

We need the sum $S = \alpha^2 + \beta^2$ and product $P = \alpha^2 \beta^2$.

$S = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 4 = 21$.

$P = (\alpha\beta)^2 = 4$.

The required equation is $x^2 - Sx + P = 0$I.e., $x^2 - 21x + 4 = 0$.

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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 $x$.

Example

Solve $3^{2x} - 4 \cdot 3^x - 5 = 0$.

Let $u = 3^x$ ($u > 0$). The equation becomes:

$u^2 - 4u - 5 = 0$

$(u - 5)(u + 1) = 0$

$u = 5$ or $u = -1$. Since $u > 0$Reject $u = -1$.

$3^x = 5 \implies x = \log_3 5 = \frac◆LB◆\ln 5◆RB◆◆LB◆\ln 3◆RB◆$.

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 $[-1, 1]$.

Example

Solve $2\cos^2\theta - \cos\theta - 1 = 0$ for $0 \leq \theta \leq 2\pi$.

Let $u = \cos\theta$ with $-1 \leq u \leq 1$:

$2u^2 - u - 1 = 0 \implies (2u + 1)(u - 1) = 0$

$u = -\frac{1}{2}$ or $u = 1$. Both lie in $[-1, 1]$.

$\cos\theta = -\frac{1}{2} \implies \theta = \frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆, \frac◆LB◆4\pi◆RB◆◆LB◆3◆RB◆$.

$\cos\theta = 1 \implies \theta = 0, 2\pi$.

Solution: $\theta \in \left\{0, \frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆, \frac◆LB◆4\pi◆RB◆◆LB◆3◆RB◆, 2\pi\right\}$.

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.

:::caution Squaring both sides of an equation is not reversible — it can introduce extraneous Solutions. You must always substitute every candidate solution back into the original equation To verify it. :::

Example

Solve $\sqrt{2x + 1} = x - 1$.

First, note the domain restrictions: $2x + 1 \geq 0$ (so $x \geq -1/2$) and $x - 1 \geq 0$ (so $x \geq 1$) since a square root is non-negative.

Squaring both sides: $2x + 1 = (x - 1)^2 = x^2 - 2x + 1$.

$x^2 - 4x = 0 \implies x(x - 4) = 0 \implies x = 0 \mathrm{ or } x = 4$

Check $x = 0$: $\sqrt{1} = -1$? No, $1 \neq -1$. Reject (also fails $x \geq 1$).

Check $x = 4$: $\sqrt{9} = 3$? Yes, $3 = 3$. Accept.

Solution: $x = 4$.

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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.

Solution

$$\begin{aligned} 3x^2 - 12x + 7 &= 3(x^2 - 4x) + 7 \\ &= 3\left[(x - 2)^2 - 4\right] + 7 \\ &= 3(x - 2)^2 - 12 + 7 \\ &= 3(x - 2)^2 - 5 \end{aligned}$$

Minimum value is $-5$Occurring at $x = 2$.

If you get this wrong, revise: Completing the square

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

Solution

$$\begin{aligned} X &= \frac◆LB◆-5 \pm \sqrt{25 + 32}◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆-5 \pm \sqrt{57}◆RB◆◆LB◆4◆RB◆ \end{aligned}$$

If you get this wrong, revise: Quadratic formula

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

Solution

We need $\Delta \geq 0$:

$$\begin{aligned} 36 - 16k &\geq 0 \\ K &\leq \frac{36}{16} = \frac{9}{4} \end{aligned}$$

Note: $k \neq 0$ (otherwise it’s not quadratic). If $k = 0$The equation $-6x + 4 = 0$ still has a Real root, so the condition is $k \leq \frac{9}{4}$ with $k$ real (including $k = 0$Which gives a Linear equation).

If you get this wrong, revise: Discriminant

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

Solution

Factorise: $(x - 5)(x + 2) > 0$.

The parabola opens upwards. It is positive outside the roots:

$x < -2 \quad \mathrm{or} \quad x > 5$

In set notation: $x \in (-\infty, -2) \cup (5, \infty)$.

If you get this wrong, revise: Quadratic inequalities

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

Solution

Substitute $y = x - 1$ into $x^2 + (x-1)^2 = 13$:

$$\begin{aligned} X^2 + x^2 - 2x + 1 &= 13 \\ 2x^2 - 2x - 12 &= 0 \\ X^2 - x - 6 &= 0 \\ (x - 3)(x + 2) &= 0 \end{aligned}$$

$x = 3$: $y = 2$. Point: $(3, 2)$.

$x = -2$: $y = -3$. Point: $(-2, -3)$.

If you get this wrong, revise: Simultaneous equations

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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 $r$.

Solution

In completed square form: $f(x) = p(x - 2)^2 - 5$.

Since $f(0) = 7$:

$$\begin{aligned} P(0 - 2)^2 - 5 &= 7 \\ 4p &= 12 \\ P &= 3 \end{aligned}$$

So $f(x) = 3(x - 2)^2 - 5 = 3(x^2 - 4x + 4) - 5 = 3x^2 - 12x + 7$.

Therefore p = 3$$q = -12$$r = 7.

If you get this wrong, revise: Completing the square

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

Solution

Completing the square:

$x^2 + 4x + 9 = (x + 2)^2 + 5$

Since $(x+2)^2 \geq 0$ for all $x$We have $(x+2)^2 + 5 \geq 5 > 0$ for all $x$. $\blacksquare$

Alternatively: $\Delta = 16 - 36 = -20 < 0$And since the coefficient of $x^2$ is positive, the Parabola is always above the $x$-axis.

If you get this wrong, revise: Discriminant

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

Solution

By polynomial division:

$2x^2 \div x = 2x$. Multiply: $2x(x-3) = 2x^2 - 6x$. Subtract: $-2x + 5$.

$-2x \div x = -2$. Multiply: $-2(x-3) = -2x + 6$. Subtract: $-1$.

So $\frac{2x^2 - 8x + 5}{x - 3} = 2x - 2 - \frac{1}{x - 3}$.

Therefore A = 2$$B = -2$$C = -1.

If you get this wrong, revise: Polynomial division

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

Solution

At $x = 1$: $y = 1 - 4 + 3 = 0$. The point is $(1, 0)$.

The gradient is $\frac{dy}{dx} = 2x - 4$. At $x = 1$: $\frac{dy}{dx} = -2$.

Using $y - y_1 = m(x - x_1)$:

$y - 0 = -2(x - 1) \implies y = -2x + 2$

If you get this wrong, revise: Differentiation

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

Solution

$$\begin{aligned} \Delta &= (2k)^2 - 4(1)(k^2 + 1) \\ &= 4k^2 - 4k^2 - 4 \\ &= -4 < 0 \end{aligned}$$

Since $\Delta = -4 < 0$ for all $k \in \mathbb{R}$There are no real roots. $\blacksquare$

If you get this wrong, revise: Discriminant

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

Solution

$(x + 5)(x + 2) = 30$ $x^2 + 7x + 10 = 30$ $x^2 + 7x - 20 = 0$

$x = \frac◆LB◆-7 \pm \sqrt{49 + 80}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-7 \pm \sqrt{129}◆RB◆◆LB◆2◆RB◆$

Since length and width must be positive, $x > -2$So:

$x = \frac◆LB◆-7 + \sqrt{129}◆RB◆◆LB◆2◆RB◆ \approx 2.18$

If you get this wrong, revise: Quadratic formula

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

Solution

$f(x) = (x - 3)^2 + 5$

Since $(x - 3)^2 \geq 0$We have $f(x) \geq 5$. Range: $f(x) \in [5, \infty)$.

For $f(x) \leq 10$:

$(x - 3)^2 + 5 \leq 10 \implies (x - 3)^2 \leq 5 \implies -\sqrt{5} \leq x - 3 \leq \sqrt{5}$

$3 - \sqrt{5} \leq x \leq 3 + \sqrt{5}$

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

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

Solution

Let $z = x^2$ (so $z \geq 0$). Then $z^2 - 13z + 36 = 0$.

$(z - 4)(z - 9) = 0$

$z = 4$ or $z = 9$.

$x^2 = 4 \implies x = \pm 2$And $x^2 = 9 \implies x = \pm 3$.

Solution: $x \in \{-3, -2, 2, 3\}$.

If you get this wrong, revise: Quadratics in disguise

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

Solution

From Vieta’s formulas: $\alpha + \beta = \frac{5}{3}$ and $\alpha\beta = \frac{1}{3}$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{9} - \frac{2}{3} = \frac{25}{9} - \frac{6}{9} = \frac{19}{9}$.

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \frac{125}{27} - 3 \cdot \frac{1}{3} \cdot \frac{5}{3} = \frac{125}{27} - \frac{5}{3} = \frac{125}{27} - \frac{45}{27} = \frac{80}{27}$.

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

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

Solution

Let $u = 4^x$ ($u > 0$). Then $u^2 - 5u + 4 = 0$.

$(u - 1)(u - 4) = 0$

$u = 1$ or $u = 4$.

$4^x = 1 \implies x = 0$And $4^x = 4 \implies x = 1$.

Solution: $x = 0$ or $x = 1$.

If you get this wrong, revise: Applying quadratic theory

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Problem 16. The curve $y = (k - 2)x^2 + 3x + k - 4$ is tangent to the $x$-axis. Find the Possible values of $k$.

Solution

For the curve to be tangent to the $x$-axis, we need $\Delta = 0$:

$\Delta = 9 - 4(k - 2)(k - 4) = 0$

$9 - 4(k^2 - 6k + 8) = 0$

$9 - 4k^2 + 24k - 32 = 0$

$4k^2 - 24k + 23 = 0$

$k = \frac◆LB◆24 \pm \sqrt{576 - 368}◆RB◆◆LB◆8◆RB◆ = \frac◆LB◆24 \pm \sqrt{208}◆RB◆◆LB◆8◆RB◆ = \frac◆LB◆24 \pm 4\sqrt{13}◆RB◆◆LB◆8◆RB◆ = \frac◆LB◆6 \pm \sqrt{13}◆RB◆◆LB◆2◆RB◆$

We also require $k \neq 2$ (otherwise the equation is linear, not quadratic). Since $\frac◆LB◆6 \pm \sqrt{13}◆RB◆◆LB◆2◆RB◆ \neq 2$ (because $6 \pm \sqrt{13} \neq 4$), both values are valid.

If you get this wrong, revise: Discriminant and graph shape

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Problem 17. Solve the inequality $x^2 + 2x - 8 \leq 0$ using a sign chart, giving your answer in Set notation.

Solution

Factorise: $(x + 4)(x - 2) \leq 0$.

Critical values: $x = -4$ and $x = 2$.

Sign chart:

Interval	$x + 4$	$x - 2$	Product
$x < -4$	$-$	$-$	$+$
$-4 < x < 2$	$+$	$-$	$-$
$x > 2$	$+$	$+$	$+$

The product is $\leq 0$ for $-4 \leq x \leq 2$.

Solution: $x \in [-4, 2]$.

If you get this wrong, revise: Quadratic inequalities

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Problem 18. Solve $2\sin^2\theta + 3\sin\theta - 2 = 0$ for $0 \leq \theta < 2\pi$.

Solution

Let $u = \sin\theta$ with $-1 \leq u \leq 1$:

$2u^2 + 3u - 2 = 0 \implies (2u - 1)(u + 2) = 0$

$u = \frac{1}{2}$ or $u = -2$.

Since $-1 \leq \sin\theta \leq 1$We reject $u = -2$.

$\sin\theta = \frac{1}{2} \implies \theta = \frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆$ or $\theta = \frac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆$.

Solution: $\theta \in \left\{\frac◆LB◆\pi◆RB◆◆LB◆6◆RB◆, \frac◆LB◆5\pi◆RB◆◆LB◆6◆RB◆\right\}$.

If you get this wrong, revise: Applying quadratic theory

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:::tip Diagnostic Test Ready to test your understanding of Quadratics? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Quadratics with other pure mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Common Pitfalls

1. Losing marks by not showing sufficient working — always write out each step, especially in proof questions.

2. Incorrectly applying integration by parts by choosing $u$ and $\frac{dv}{dx}$ the wrong way around.

3. Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.

4. Rounding too early in multi-step calculations — carry full precision through and round only the final answer.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

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