2. Equations and Inequalities

🧠 What This Chapter Is Really About

In Chapter 1, you learned to read quadratic functions — their shape, position, and turning points. Now it’s time to solve them. This chapter is where you crack open equations to find exact roots, tackle systems where two relationships collide, and map out the ranges of values where an expression stays positive or negative. Think of this chapter as giving you the algebraic engine that powers everything from projectile analysis to engineering design constraints.

Why should you care? Every time an engineer sizes a beam, a game designer plots a trajectory, or a financial analyst models break-even points, they are solving equations and interpreting inequalities exactly as you will in this chapter. These aren’t abstract exercises — they’re the real tools professionals reach for daily.

A note on E-Math vs A-Math: If you’ve covered completing the square and the quadratic formula in E-Math, great — you have a head start. But A-Math goes much deeper. Here, you must derive the quadratic formula from completing the square, handle non-linear simultaneous equations (line–parabola, line–circle), and solve quadratic inequalities with rigorous sign-analysis. None of these appear in the E-Math syllabus, so don’t assume you can coast on what you already know!

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🔹 2.1 Solving Quadratic Equations by Completing the Square

🧠 Key Concepts

The core idea behind completing the square is beautifully simple: rewrite $a{x}^2+bx+c$ in the form $a(x-h{)}^2+k$ so that $x$ can be isolated by taking a square root. Once you see the expression as a perfect square plus (or minus) a constant, solving becomes almost mechanical.

Here’s the general completed-square result you should know:

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

When $a=1$, the process simplifies nicely:

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

The idea is: take half the coefficient of $x$, square it, then add and subtract that value to create a perfect square trinomial.

When $a\ne 1$, your first step is always to factor out $a$ from the $x^2$ and $x$ terms before completing the square. Skipping this step is one of the most common sources of messy algebra — don’t fall into that trap!

Why does this matter beyond just solving equations? Two big reasons:

1. Completing the square is the proof method behind the Quadratic Formula. A-Math expects you to reproduce this derivation — it’s an exam favourite.
2. Link back to Chapter 1: Completing the square also gives you the vertex $(h,k)$ of the parabola $y=a{x}^2+bx+c$. So it’s doing double duty.

One more thing: equations with irrational roots (surd answers) are naturally handled by this method. When a question says “leave answers in exact form,” completing the square is your go-to approach.

🏷️ Definitions & Terminology

• Perfect square trinomial: An expression of the form $(x+p{)}^2=x^2+2px+p^2$. Recognising this pattern is the heart of the technique.
• Completing the square: The algebraic technique of adding and subtracting ${\left(\frac{b}{2a}\right)}^2$ to create a perfect square trinomial inside $a{x}^2+bx+c$.
• Vertex form: $a(x-h{)}^2+k$, where $(h,k)$ is the turning point of the parabola. This is the “finished product” of completing the square.
• Surd form: An answer left in exact radical notation, e.g. $3+\sqrt{5}$, rather than a decimal approximation.

Worked Examples

Easy

Basic completing the square with $a=1$ and integer roots Question: Solve $x^2+6x+5=0$ by completing the square. Let’s start with a friendly one where the roots come out as whole numbers, so you can focus on the process. Step 1 (Halve the coefficient of $x$ and square it): The coefficient of $x$ is $6$. Half of $6$ is $3$, and $3^2=9$. Step 2 (Rewrite as a perfect square): Add and subtract $9$: $x^2+6x+9-9+5=0$ $(x+3{)}^2-4=0$ Step 3 (Isolate the square and take the square root): $\begin{array}{rl}(x+3{)}^2& =4\\ x+3& =\pm 2\end{array}$ Step 4 (Solve for $x$): $x=-3+2=-1\text{or}x=-3-2=-5$ See how clean that is? The roots are $x=-1$ and $x=-5$. You could verify by factorising: $x^2+6x+5=(x+1)(x+5)$. Same answer!

Expressing in vertex form Question: Express $x^2-4x+7$ in the form $(x-h{)}^2+k$ and state the minimum value. Step 1 (Halve the coefficient of $x$ and square it): The coefficient of $x$ is $-4$. Half of $-4$ is $-2$, and $(-2{)}^2=4$. Step 2 (Rewrite): $x^2-4x+4-4+7=(x-2{)}^2+3$ So in vertex form: $(x-2{)}^2+3$. Step 3 (State the minimum): Since $(x-2{)}^2\ge 0$ for all real $x$, the smallest value of $(x-2{)}^2+3$ is $0+3=3$, occurring when $x=2$. The minimum value is $3$. Notice how completing the square immediately tells you the turning point is $(2,3)$ — exactly what you learned in Chapter 1!

Medium

Completing the square with $a\ne 1$ and surd roots Question: Solve $2x^2-8x-3=0$ by completing the square, leaving answers in surd form. Don’t be intimidated by the leading coefficient of $2$ — just factor it out first! Step 1 (Factor out $a=2$ from the $x^2$ and $x$ terms): $2\left(x^2-4x\right)-3=0$ Step 2 (Complete the square inside the bracket): Half of $-4$ is $-2$, and $(-2{)}^2=4$. $2\left(x^2-4x+4-4\right)-3=0$ $2\left((x-2{)}^2-4\right)-3=0$ Step 3 (Expand and simplify): $2(x-2{)}^2-8-3=0$ $2(x-2{)}^2=11$ Step 4 (Isolate and take the square root): $\begin{array}{rl}(x-2{)}^2& =\frac{11}{2}\\ x-2& =\pm \sqrt{\frac{11}{2}}\\ x& =2\pm \sqrt{\frac{11}{2}}\end{array}$ To tidy up the surd (rationalise the denominator): $x=2\pm \frac{\sqrt{22}}{2}$ Both forms are acceptable as “surd form.” The key step was factoring out the $2$ at the very beginning — if you skip that, you’ll end up in a fraction mess.

Derivation of the Quadratic Formula (Exam favourite!) Question: Starting from $a{x}^2+bx+c=0$ (where $a\ne 0$), derive $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ using completing the square. This is one of the most commonly tested derivations in A-Math. Make sure you can reproduce every line. Step 1 (Divide through by $a$): $x^2+\frac{b}{a}x+\frac{c}{a}=0$ Step 2 (Move the constant term to the right): $x^2+\frac{b}{a}x=-\frac{c}{a}$ Step 3 (Complete the square on the left): Half of $\frac{b}{a}$ is $\frac{b}{2a}$, and ${\left(\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}$. Add this to both sides: $x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$ ${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-\frac{c}{a}$ Step 4 (Combine the right side over a common denominator): ${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2}$ Step 5 (Take the square root of both sides): $x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$ Step 6 (Isolate $x$): $x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ And there it is — the Quadratic Formula, born from completing the square. Every time you use the formula, this is what’s happening behind the scenes.

Hard

Completing the square in a word problem Question: The profit $P=-3x^2+18x-5$ (in thousands of dollars), where $x$ is the number of units produced (in hundreds). Find the value of $x$ that maximises profit and the maximum profit, using completing the square. Step 1 (Factor out $a=-3$ from the $x^2$ and $x$ terms): $P=-3(x^2-6x)-5$ Step 2 (Complete the square inside the bracket): Half of $-6$ is $-3$, and $(-3{)}^2=9$. $P=-3(x^2-6x+9-9)-5$ $P=-3\left((x-3{)}^2-9\right)-5$ Step 3 (Expand): $P=-3(x-3{)}^2+27-5$ $P=-3(x-3{)}^2+22$ Step 4 (Interpret): Since $-3(x-3{)}^2\le 0$ for all real $x$ (a negative number times a square is always $\le 0$), the maximum value of $P$ occurs when $(x-3{)}^2=0$, i.e. when $x=3$. Maximum profit $=22$ thousand dollars, when $x=3$ (i.e., 300 units produced). Notice how the negative leading coefficient means the parabola opens downward — so it has a maximum, not a minimum. This connects directly to Chapter 1!

🌐 Real-world Link

GPS receivers solve systems of quadratic equations (involving distances squared) to pinpoint your location. Completing the square is the algebraic backbone of simplifying those distance equations — so this technique is literally helping satellites find you right now.

🤔 Check Yourself

1. Complete the square for $x^2+10x+18$ and hence solve $x^2+10x+18=0$, giving answers in surd form.
2. Without using the quadratic formula, explain why completing the square on $a{x}^2+bx+c=0$ must produce the quadratic formula.

⚠️ Common Mistakes

• Have you ever added ${\left(\frac{b}{2}\right)}^2$ to one side of an equation but not the other? Why does this break the equation, and how should you handle it when completing the square inside an expression (rather than an equation)?
• Did you try completing the square on $3x^2+12x$ without dividing by $3$ first? What goes wrong, and what should you always do when $a\ne 1$?
• If you reach $(x+3{)}^2=-2$, should you take the square root? What does a negative number on the right tell you about real solutions?
• For $x^2-10x$, the perfect square is $(x-5{)}^2$, not $(x+5{)}^2$. Why does the sign inside the bracket always match the sign of half the coefficient of $x$?

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🔹 2.2 Solving Quadratic Equations Using the Quadratic Formula

🧠 Key Concepts

The Quadratic Formula is your universal key — it solves any equation of the form $a{x}^2+bx+c=0$ (provided $a\ne 0$):

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

But the real power of this section isn’t just plugging numbers into a formula. It’s learning to use the discriminant $D=b^2-4ac$ as a diagnostic tool that tells you the nature of roots before you even solve:

• $D>0$: two distinct real roots
• $D=0$: two equal (repeated) real roots — the parabola just touches the $x$-axis
• $D<0$: no real roots — the parabola doesn’t cross the $x$-axis at all

Here’s a bonus insight: if $D$ is a perfect square (like $25$, $49$, $100$), the roots are rational. If $D$ is positive but not a perfect square (like $5$, $13$, $41$), the roots are irrational (surds).

A-Math frequently asks condition problems: “Find the value(s) of $k$ such that the equation has equal roots / no real roots / two distinct real roots.” For these, you go straight to the discriminant — set $D=0$, $D<0$, or $D>0$ and solve for $k$.

A few practical reminders:

• When the question says “correct to 2 decimal places” or “3 significant figures,” use the formula and evaluate with a calculator.
• When it says “exact form,” leave your answer in surd form.
• Always rearrange to $a{x}^2+bx+c=0$ before identifying $a$, $b$, $c$. Getting this wrong is a surprisingly common way to throw away marks.

🏷️ Definitions & Terminology

• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ — the universal root-finding formula for quadratics.
• Discriminant ($D$): $D=b^2-4ac$; it determines the number and type of real roots without requiring you to solve.
• Nature of roots: A classification — 2 distinct real, 2 equal real, or no real roots — based on the sign of $D$.
• Repeated / equal / double root: When $D=0$, the equation has one root $x=-\frac{b}{2a}$ with multiplicity 2. You’ll sometimes hear it called a “double root.”
• Condition problem: A problem where a parameter (e.g. $k$) must satisfy $D>0$, $D=0$, or $D<0$.

📐 Understanding the Parameters

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

• $a$ is the coefficient of $x^2$ — it must not be zero (otherwise you don’t have a quadratic!)
• $b$ is the coefficient of $x$
• $c$ is the constant term

The $\pm$ symbol means you compute two values: one with $+$ and one with $-$. These give the two roots.

There’s also a pair of relationships you’ll use frequently alongside the formula — the Sum and Product of Roots:

$x_1+x_2=-\frac{b}{a},x_1\cdot x_2=\frac{c}{a}$

These let you find expressions involving the roots without actually solving for each root individually — a powerful shortcut.

Worked Examples

Easy

Direct application with distinct real roots Question: Solve $3x^2+x-2=0$ using the quadratic formula. Step 1 (Identify $a$, $b$, $c$): Here $a=3$, $b=1$, $c=-2$. Step 2 (Compute the discriminant): $D=b^2-4ac=(1{)}^2-4(3)(-2)=1+24=25$ Since $D=25>0$ and $25$ is a perfect square, we expect two distinct rational roots. Step 3 (Apply the formula): $x=\frac{-1\pm \sqrt{25}}{2(3)}=\frac{-1\pm 5}{6}$ $x=\frac{-1+5}{6}=\frac{4}{6}=\frac{2}{3}\text{or}x=\frac{-1-5}{6}=\frac{-6}{6}=-1$ The roots are $x=\frac{2}{3}$ and $x=-1$.

Equal roots — verifying $D=0$ Question: Solve $x^2-4x+4=0$. Confirm the discriminant is zero and state the repeated root. Step 1 (Identify $a$, $b$, $c$): $a=1$, $b=-4$, $c=4$. Step 2 (Compute the discriminant): $D=(-4{)}^2-4(1)(4)=16-16=0$ Since $D=0$, there is a repeated root. Step 3 (Apply the formula): $x=\frac{-(-4)\pm \sqrt{0}}{2(1)}=\frac{4}{2}=2$ The repeated root is $x=2$. Geometrically, the parabola $y=x^2-4x+4$ just touches the $x$-axis at $x=2$ — it doesn’t cross it.

Medium

Irrational roots, answer to 3 significant figures Question: Solve $2x^2+5x-1=0$, giving answers correct to 3 significant figures. Step 1 (Identify $a$, $b$, $c$): $a=2$, $b=5$, $c=-1$. Step 2 (Compute the discriminant): $D=5^2-4(2)(-1)=25+8=33$ $D=33>0$ but $33$ is not a perfect square, so the roots will be irrational. That’s fine — the question wants decimal answers. Step 3 (Apply the formula): $x=\frac{-5\pm \sqrt{33}}{2(2)}=\frac{-5\pm \sqrt{33}}{4}$ $x=\frac{-5+\sqrt{33}}{4}=\frac{-5+5.7445\dots }{4}=\frac{0.7445\dots }{4}=0.186\text{ (3 s.f.)}$ $x=\frac{-5-\sqrt{33}}{4}=\frac{-5-5.7445\dots }{4}=\frac{-10.7445\dots }{4}=-2.69\text{ (3 s.f.)}$ The roots are $x=0.186$ and $x=-2.69$ (to 3 s.f.).

Discriminant condition problem Question: The equation $x^2+kx+9=0$ has equal roots. Find the value(s) of $k$. Step 1 (Identify $a$, $b$, $c$ in terms of $k$): $a=1$, $b=k$, $c=9$. Step 2 (Set the discriminant condition): For equal roots, $D=0$. $D=k^2-4(1)(9)=0$ $k^2-36=0$ Step 3 (Solve for $k$): $k^2=36$ $k=\pm 6$ Both values are valid! When $k=6$, the equation is $x^2+6x+9=(x+3{)}^2=0$, giving $x=-3$. When $k=-6$, it’s $x^2-6x+9=(x-3{)}^2=0$, giving $x=3$. Both are legitimate equal-root equations.

Hard

Discriminant condition with parameter in two coefficients Question: The equation $(k+1)x^2+2kx+(k-3)=0$ has no real roots. Find the range of values of $k$. Step 1 (Identify $a$, $b$, $c$): $a=k+1$, $b=2k$, $c=k-3$. But $a\ne 0$, so $k\ne -1$. Step 2 (Set $D<0$ for no real roots): $\begin{array}{rl}D& =(2k{)}^2-4(k+1)(k-3)=4k^2-4(k^2-2k-3)=8k+12\end{array}$ Step 3 (Solve $D<0$): $8k+12<0⟹k<-\frac{3}{2}$ Step 4 (Combine with $k\ne -1$): $k<-\frac{3}{2}$ already excludes $-1$. The range is $k<-\frac{3}{2}$.

Sum and product of roots Question: If $\alpha$ and $\beta$ are roots of $2x^2-7x+3=0$, find the value of $\frac{1}{\alpha }+\frac{1}{\beta }$ without finding $\alpha$ and $\beta$ individually. Step 1 (Sum and product of roots): $\alpha +\beta =\frac{7}{2}$, $\alpha \beta =\frac{3}{2}$. Step 2 (Rewrite): $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }$. Step 3 (Substitute): $\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{2}\times \frac{2}{3}=\frac{7}{3}$.

🌐 Real-world Link

Pharmacologists use the quadratic formula to calculate drug dosage thresholds. The concentration of a drug in the bloodstream is often modelled as a quadratic function of time, and the formula identifies the exact moments when the concentration reaches a critical level — too much is toxic, too little is ineffective.

🤔 Check Yourself

1. Solve $5x^2-3x-1=0$, giving your answers in the form $\frac{p\pm \sqrt{q}}{r}$.
2. For what values of $m$ does $m{x}^2+4x+m=0$ have two distinct real roots? (Don’t forget to check $m\ne 0$.)

⚠️ Common Mistakes

• Have you accidentally written $b=6x$ instead of $b=6$? Remember: $a$, $b$, and $c$ must be constants, not expressions containing $x$. If you see $x$ in your values of $a$, $b$, or $c$, something has gone wrong.
• Where exactly does $2a$ go in the formula? It’s $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ — the entire numerator is divided by $2a$. Writing $\frac{-b\pm \sqrt{b^2-4ac}}{2}\times a$ is a different (and wrong!) expression.
• Did you set $D>0$ for “no real roots”? That’s backwards! No real roots means $D<0$. A quick mnemonic: “no roots, negative discriminant.”
• Did you forget to rearrange $x^2=5x-3$ into standard form? Move everything to one side first: $x^2-5x+3=0$. The values of $b$ and $c$ change completely if you skip this step.
• For condition problems, you might get two values of $k$ from $D=0$. Should you reject one? Not necessarily — always check both. Often both values are valid.

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🔹 2.3 Solving Linear and Non-linear Simultaneous Equations

🧠 Key Concepts

You’ve solved linear–linear systems before in E-Math using substitution or elimination. That’s a great foundation. But in A-Math, things get more interesting: you’ll solve linear–quadratic systems — a straight line paired with a parabola or a circle.

The method is substitution: express one variable from the linear equation, plug it into the non-linear equation, and solve the resulting quadratic. Here’s the crucial insight:

The resulting quadratic can have $0$, $1$, or $2$ solutions — and this corresponds geometrically to the line:

• Missing the curve entirely ($D<0$: no intersection)
• Being tangent to the curve ($D=0$: exactly one point of contact)
• Intersecting the curve at two points ($D>0$: two intersection points)

So the discriminant of the resulting quadratic tells you the number of intersection points — a beautiful connection between algebra and geometry.

After finding $x$-values, always substitute back into the linear equation (not the quadratic) to find corresponding $y$-values. Why? Because the quadratic can introduce extraneous solutions (values that satisfy the quadratic but not the original system). The linear equation is safe.

Watch out for systems involving $x^2+y^2=r^2$ — equations of circles. These are very common in A-Math and don’t appear in E-Math.

🏷️ Definitions & Terminology

• Simultaneous equations: Two or more equations that share common unknowns; solved together to find values satisfying all equations at once.
• Linear equation: An equation of degree 1, e.g. $y=2x+3$.
• Non-linear equation: An equation of degree $\ge 2$, e.g. $y=x^2-4x+1$ or $x^2+y^2=25$.
• Substitution method: Rearrange one equation for one variable and substitute into the other. This is your primary weapon for linear–quadratic systems.
• Elimination method: Add or subtract equations to remove one variable. This works mainly for linear–linear systems.
• Tangent to a curve: A line that touches the curve at exactly one point — corresponds to $D=0$ in the resulting quadratic.
• Extraneous solution: A “solution” introduced by algebraic manipulation that does not satisfy the original system. Always verify!

🧠 Key Concept

For the system $y=mx+c$ and $y=a{x}^2+bx+d$, substitute the linear equation into the quadratic:

$a{x}^2+bx+d=mx+c⟹a{x}^2+(b-m)x+(d-c)=0$

The number of intersection points is determined by the discriminant of this resulting quadratic:

$D=(b-m{)}^2-4a(d-c)$

• $D>0$: 2 points of intersection
• $D=0$: 1 point (tangency)
• $D<0$: no intersection

For a line $y=mx+c$ and a circle $x^2+y^2=r^2$, substitute to get:

$x^2+(mx+c{)}^2=r^2⟹(1+m^2)x^2+2mcx+(c^2-r^2)=0$

Same discriminant analysis applies!

Worked Examples

Easy

Linear–linear system (revision) Question: Solve $2x+3y=12$ and $x-y=1$. Step 1 (Express $x$ from the simpler equation): From $x-y=1$: $x=y+1$ Step 2 (Substitute into the other equation): $2(y+1)+3y=12⟹2y+2+3y=12⟹5y=10⟹y=2$ Step 3 (Find $x$): $x=y+1=2+1=3$ The solution is $x=3$, $y=2$.

Medium

Linear–quadratic: line meets parabola Question: Solve $y=x+2$ and $y=x^2-2x+3$. Step 1 (Substitute the linear equation into the quadratic): $x+2=x^2-2x+3⟹x^2-3x+1=0$ Step 2 (Solve the resulting quadratic): $D=9-4=5$, so $x=\frac{3\pm \sqrt{5}}{2}$. Step 3 (Find corresponding $y$ using $y=x+2$): $y=\frac{3\pm \sqrt{5}}{2}+2=\frac{7\pm \sqrt{5}}{2}$ Intersection points: $(\frac{3+\sqrt{5}}{2},\frac{7+\sqrt{5}}{2})$ and $(\frac{3-\sqrt{5}}{2},\frac{7-\sqrt{5}}{2})$.

Linear–circle system Question: Solve $y=x+2$ and $x^2+y^2=8$. Substitute: $x^2+(x+2{)}^2=8⟹2x^2+4x-4=0⟹x^2+2x-2=0$. $D=4+8=12$, so $x=-1\pm \sqrt{3}$; $y=x+2=1\pm \sqrt{3}$.

🌐 Real-world Link

Navigation satellites determine your position by solving a system of non-linear equations: each satellite provides a circle (or sphere in 3D) of possible locations, and the intersection of these curves pins down your exact coordinates. The algebra behind GPS is essentially the same substitution method you’ve just learned.

🤔 Check Yourself

1. Solve the system $y=2x-1$ and $x^2+y^2=10$, leaving answers in surd form where necessary.
2. Explain geometrically what it means when the discriminant of the resulting quadratic equals zero.

⚠️ Common Mistakes

• Did you substitute back into the quadratic equation and get extra answers? Use the linear equation for back-substitution.
• You found $x$ but forgot the corresponding $y$. An ordered pair requires both values!
• Trying elimination on $x^2+y^2=25$ and $y=x+1$? Substitution is cleaner.
• Is $D=0$ “no solution” or “exactly one solution”? It’s exactly one — tangency.
• Squaring both sides of $y=x+1$ can introduce extraneous solutions. Direct substitution is safer.

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🔹 2.4 Solving Quadratic Inequalities

🧠 Key Concepts

Here’s the big idea: to solve $a{x}^2+bx+c>0$ (or $<$, $\ge$, $\le$), find the roots of $a{x}^2+bx+c=0$. These roots divide the number line into intervals, and within each interval the quadratic has a consistent sign.

If $a>0$ (U-shape):

• Expression is negative between the roots and positive outside
• $a{x}^2+bx+c>0⟺x<\alpha \text{ or }x>\beta$
• $a{x}^2+bx+c<0⟺\alpha <x<\beta$

If $a<0$ (n-shape):

• Expression is positive between the roots and negative outside
• $a{x}^2+bx+c>0⟺\alpha <x<\beta$
• $a{x}^2+bx+c<0⟺x<\alpha \text{ or }x>\beta$

For $\ge$ and $\le$, include the endpoints. Special cases:

• $D=0$: one repeated root — the expression is zero at that point and one sign elsewhere.
• $D<0$: no real roots — the expression has the same sign for all $x$.

Remember: multiplying/dividing an inequality by a negative flips the sign.

🏷️ Definitions & Terminology

• Critical values: roots of $a{x}^2+bx+c=0$, dividing the number line.
• Test interval: sub-interval between critical values.
• Sign diagram: shows $+$/$-$ in each interval.
• Interval notation: e.g. $(-\infty ,2)\cup (3,\infty )$.
• Set-builder notation: $\{x:x<2\text{ or }x>3\}$.

Worked Examples

Easy

Standard $>0$ with factorable quadratic Question: Solve $x^2-5x+6>0$. Roots: $x=2,3$. U-shape ($a=1>0$), positive outside: $x<2$ or $x>3$. Standard $\le 0$ with factorable quadratic Question: Solve $x^2-x-12\le 0$. Roots: $x=-3,4$. U-shape, on/below between: $-3\le x\le 4$.

Medium

Negative leading coefficient with $\ge$ Solve $-2x^2+8x+4\ge 0$. Multiply by $-1$ (flip): $2x^2-8x-4\le 0$. Divide by 2: $x^2-4x-2\le 0$. Roots $2\pm \sqrt{6}$. Solution: $2-\sqrt{6}\le x\le 2+\sqrt{6}$. Non-factorable with formula Solve $2x^2+3x-4<0$. Roots: $\frac{-3\pm \sqrt{41}}{4}$. Interval between.

Hard

Special case $D=0$ Solve $x^2-6x+9\ge 0$. $(x-3{)}^2\ge 0$ true for all real $x$. Special case $D<0$ Solve $x^2+2x+5>0$. $D=-16<0$, $a>0$, true for all real $x$. Parameter problem Find $k$ for $k{x}^2+4x+k>0\forall x$. Need $k>0$ and $D<0$: $16-4k^2<0⟹k^2>4$. Combined: $k>2$.

🌐 Real-world Link

Engineers design safety margins using quadratic inequalities. For example, stress $\sigma (x)$ on a beam is a quadratic in $x$, and ensuring $\sigma (x)\le {\sigma }_{max}$ tells exactly which sections are safe.

🤔 Check Yourself

1. Solve $3x^2-2x-8\ge 0$ and represent on a number line.
2. Without solving, explain why $x^2+4>0$ for all real $x$.

⚠️ Common Mistakes

• Sketch the U-shape: $x^2-5x+6>0$ is $x<2$ or $x>3$, not $2<x<3$.
• Negatives flip the sign — forgetting this reverses your answer.
• No real roots ($D<0$) & $a>0$ means true for all $x$, not “no solution”.
• Exclude endpoints for strict inequalities, include for $\ge$/$\le$.
• Check sign of $a$ — U-shape vs n-shape changes region signs.

✅ Quick Strategy 🎯

1. Choose your weapon wisely. Completing the square for vertex form or exact surd roots/derivations. Quadratic Formula for speed.
2. Always standardise: $a{x}^2+bx+c=0$ before identifying $a,b,c$.
3. Discriminant first: compute $D=b^2-4ac$ to predict root nature.
4. Substitute from linear in mixed systems — avoid extraneous solutions.
5. Sketch first for inequalities — a quick U/n shape wins time.
6. Flip on negatives: multiply/divide inequalities by negative → flip symbol.