Quadratics

More Than Just a Smiley Face! 😊

Introduction

1. Introduction

ax^2 + bx + c

2. Expressing Quadratics in Completed Square Form

ax^2+bx+c

A diagram showing a U-shaped parabola with its vertex labelled (h, k). The completed square form is written next to it, with arrows pointing from '-h' to the x-coordinate 'h' (highlighting the sign change) and from '+k' to the y-coordinate 'k'.

a(x+p)^2+q

Worked example

Worked Example: Finding the Vertex by Completing the Square

Let's Solve This Thing 🚀

2x^2 - 12x + 5

1
$x^2$
$2(x^2 - 6x) + 5$
2
$x^2 - 6x$
$2(x^2 - 6x + 9 - 9) + 5$
3
$x^2 - 6x + 9$
$2((x-3)^2 - 9) + 5$
4
$2 \times -9 = -18$
$2(x-3)^2 - 18 + 5$
5
$-18 + 5 = -13$
$2(x-3)^2 - 13$
6
$a(x+p)^2+q$
$Vertex = (3, -13)$

Answer

Vertex = (3, -13)

3. Using the Discriminant to Determine the Nature of Roots

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

A U-shaped parabola intersecting the x-axis at two distinct points, e.g., x=-1 and x=3.

b^2 - 4ac = 0

A U-shaped parabola with its vertex touching the x-axis at a single point, e.g., x=2.

b^2 - 4ac < 0

A U-shaped parabola floating entirely above the x-axis, not touching it at all.

b^2 - 4ac

Worked example

Worked Example: Finding Values for a Condition of Tangency

When a Line and a Curve Just Barely Touch 🤝

y = kx - 3

1
$First up, let's unpack the word 'tangent'. It's a huge clue! It means the line and the curve meet at exactly one point . To find intersection points, we always set the equations equal to each other. This will create a new quadratic, which must have exactly one repeated root.$
$kx - 3 = 2x^2 - 5x + 1$
2
$ax^2 + bx + c = 0$
$2x^2 - 5x - kx + 1 + 3 = 0 \\ 2x^2 - (5+k)x + 4 = 0$
3
$a, b,$
$b^2 - 4ac = 0$
4
$b$
$(-(5+k))^2 - 4(2)(4) = 0$
5
$k$
$(5+k)^2 - 32 = 0 \\ (5+k)^2 = 32$
6
$k$
$5+k = \pm \sqrt{32} \\ 5+k = \pm 4\sqrt{2} \\ k = -5 \pm 4\sqrt{2}$

Answer

5+k = \pm \sqrt{32} \\ 5+k = \pm 4\sqrt{2} \\ k = -5 \pm 4\sqrt{2}

4. Solving Quadratic Inequalities

ax^2 + bx + c > 0

A diagram showing a U-shaped parabola crossing the x-axis at two critical values, 'a' and 'b'. The region between 'a' and 'b' is shaded below the x-axis (for y < 0), and the regions x < a and x > b are shaded above the x-axis (for y > 0).

> 0

Worked example

Worked Example: Solving a Quadratic Inequality

Let's Crack This Code

x

1
$First up, we need to find the 'critical values'. To do this, we pretend the inequality is an equation and solve it to find the x-intercepts.$
$x^2 - x - 12 = 0$
2
$Let's factorize the quadratic to find the roots. We're looking for two numbers that multiply to -12 and add to -1.$
$(x - 4)(x + 3) = 0 \\ \text{So, the critical values are } x = 4 \text{ and } x = -3.$
3
$y = x^2 - x - 12$
$\text{Visual aid: Imagine a U-shaped curve passing through x = -3 and x = 4.}$
4
$x^2 - x - 12 > 0$
$\text{Regions identified on the sketch are where y > 0.}$
5
$Finally, we write this down as our solution. Remember to use 'or' because x can't be in both places at once. It's either in one zone or the other. This is the final answer format the examiners are looking for!$
$x < -3 \text{ or } x > 4$

Answer

x < -3 \text{ or } x > 4

5. Solving Simultaneous Linear and Quadratic Equations

y = mx + c

A graph showing a straight line intersecting a U-shaped parabola at two distinct points, labelled P and Q, with their coordinates shown.

Once you've done the substitution, you'll have a standard quadratic equation to solve. You can factorise it, use the quadratic formula - whatever your go-to move is. Solving this will give you the x-coordinate(s) of the intersection points. But wait, don't forget their partners! Each x-value has a corresponding y-value. To find them, sub your x-values back into the original linear equation (trust me, it's way easier). You might get two solutions, one solution (if the line just grazes the curve, like a tangent), or no real solutions (they completely miss each other). And that's it - you've found the exact coordinates where the line and the curve meet. 🎯

Worked example

Worked Example: Intersection of a Line and a Parabola

Let's Find Where They Meet Up ✨

y - 2x = 3

1
$y - 2x = 3$
$y = 2x + 3$
2
$2x + 3$
$2x + 3 = x^2 - 2x + 7$
3
$ax^2 + bx + c = 0$
$0 = x^2 - 4x + 4$
4
$x$
$(x-2)(x-2) = 0 \implies x = 2$
5
$x=2$
$y = 2(2) + 3 = 4 + 3 = 7$
6
$And we're done! State the final answer as a coordinate pair. This is the single point where the line and the curve touch.$
$\text{Point of intersection} = (2, 7)$

Answer

\text{Point of intersection} = (2, 7)

6. Equations Reducible to Quadratic Form

x^4

A side-by-side comparison. On the left, the equation with and highlighted. An arrow points to the right, labeled 'Let u = x²'. On the right, the transformed equation with and highlighted.

u

Worked example

Worked Example: Solving a Quartic Equation via Substitution

Unmasking the Impostor Equation 🕵️‍♀️

2x^4 - 11x^2 + 12 = 0

1
$x^4$
$(x^2)^2 = x^4$
2
$u$
$Let \ u = x^2. \ The \ equation \ becomes: \ 2u^2 - 11u + 12 = 0$
3
$u$
$2u^2 - 8u - 3u + 12 = 0 <br> 2u(u - 4) - 3(u - 4) = 0 <br> (2u - 3)(u - 4) = 0 <br> So, \ u = \frac{3}{2} \ or \ u = 4$
4
$x$
$Case \ 1: \ u = 4 \implies x^2 = 4 \implies x = \pm \sqrt{4} \implies x = \pm 2 <br> Case \ 2: \ u = \frac{3}{2} \implies x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}}$
5
$x$
$The \ solutions \ are \ x = 2, \ x = -2, \ x = \sqrt{\frac{3}{2}}, \ and \ x = -\sqrt{\frac{3}{2}}$

Answer

The \ solutions \ are \ x = 2, \ x = -2, \ x = \sqrt{\frac{3}{2}}, \ and \ x = -\sqrt{\frac{3}{2}}

7. Transforming Quadratic Expressions by Completing the Square

ax^2+bx+c

A diagram showing a square of side x, two rectangles of x by b/2, and a small missing square of (b/2) by (b/2). The image visually represents how adding the (b/2)^2 term 'completes' the larger square of side (x+b/2).

a>0

Worked example

Worked Example: Finding the Vertex and Nature of a Quadratic Function

Let's Wrestle This Quadratic Into Shape 💪

f(x) = 3x^2 - 12x + 5

1
$x^2$
$f(x) = 3(x^2 - 4x) + 5$
2
$(x^2 - 4x)$
$f(x) = 3(x^2 - 4x + 4 - 4) + 5$
3
$The first three terms inside the bracket now form a perfect square. We can factorise this. The '-4' term needs to be moved outside the main bracket, but don't forget to multiply it by the 3 we factored out in step 1. This is the most common mistake!$
$f(x) = 3(x-2)^2 - (3 \times 4) + 5$
4
$Finally, simplify the constant terms outside the bracket to get our final completed square form.$
$f(x) = 3(x-2)^2 - 12 + 5 \\ f(x) = 3(x-2)^2 - 7$
5
$a(x-h)^2+k$
$a=3, h=2, k=-7 \\ \text{Vertex: } (2, -7) \\ \text{Nature: Minimum point (since a > 0)}$

Answer

a=3, h=2, k=-7 \\ \text{Vertex: } (2, -7) \\ \text{Nature: Minimum point (since a > 0)}

8. The Discriminant and the Nature of Roots

ax^2 + bx + c = 0

A graph showing three parabolas against the x-axis. The first, labelled 'b²-4ac > 0', clearly intersects the x-axis at two distinct points. The second, 'b²-4ac = 0', has its vertex touching the x-axis at a single point. The third, 'b²-4ac < 0', is positioned entirely above the x-axis, not touching it at all.

Understanding the discriminant isn't just about solving for x; it's about analysing the structure of an equation, a core skill for any future mathematician, engineer, or economist.

Worked example

Worked Example: Determining Tangency Conditions

Solving the Mystery of the Unknown Constant 'k' 🕵️‍♀️

y = kx - 2

1
$y$
$x^2 + 5x + 7 = kx - 2$
2
$ax^2 + bx + c = 0$
$x^2 + 5x - kx + 7 + 2 = 0 \\ x^2 + (5 - k)x + 9 = 0$
3
$a$
$a = 1, \quad b = (5 - k), \quad c = 9$
4
$Here's the crucial insight. The word'tangent' implies there is exactly one point of intersection. For a quadratic equation, this means there is one repeated real root. The condition for this is that the discriminant must be equal to zero.$
$b^2 - 4ac = 0$
5
$a$
$(5 - k)^2 - 4(1)(9) = 0 \\ (5 - k)^2 - 36 = 0$
6
$k$
$(5 - k)^2 = 36 \\ 5 - k = \pm\sqrt{36} \\ 5 - k = 6 \quad \text{or} \quad 5 - k = -6 \\ k = -1 \quad \text{or} \quad k = 11$

Answer

(5 - k)^2 = 36 \\ 5 - k = \pm\sqrt{36} \\ 5 - k = 6 \quad \text{or} \quad 5 - k = -6 \\ k = -1 \quad \text{or} \quad k = 11

9. Solving Quadratic Inequalities

ax^2 + bx + c = 0

A sketch of a U-shaped parabola crossing the x-axis at two critical values, and . The region where the curve is above the axis () is shown to correspond to and . The region below () corresponds to .

>

Worked example

Worked Example: Solving a Quadratic Inequality

Let's put the theory into practice! 💪

x

1
$2x^2 + 5x - 12 = 0$
$(2x - 3)(x + 4) = 0$
2
$y = 2x^2 + 5x - 12$
$[IMAGE_PLACEHOLDER_2: A simple sketch of a U-shaped parabola. It is shown crossing the x-axis at x=-4 on the left and x=1.5 on the right.]$
3
$2x^2 + 5x - 12 < 0$
4
$Finally, we write down our solution as a single inequality. Since the solution is the region between the critical values, we express it as a compound inequality. This is the final set of values that satisfies the condition.$
$-4 < x < \frac{3}{2}$

Answer

-4 < x < \frac{3}{2}

10. Solving Simultaneous Linear and Quadratic Equations

y = mx + c

A graph showing three scenarios: a line intersecting a parabola at two distinct points (2 real roots), a line touching the parabola at one point, i.e., a tangent (1 repeated root), and a line missing the parabola completely (no real roots).

So, to recap the process: 1. Isolate a variable in the linear equation. 2. Substitute into the quadratic equation. 3. Solve the resulting quadratic. 4. Find the corresponding coordinates. Master this, and you've got a key skill locked down for your exams and beyond.

Worked example

Worked Example: Finding Points of Intersection

Let's Get These Coordinates! 🚀

y = x + 1

1
$y = ...$
$x + 1 = x^2 - 4x + 5$
2
$ax^2 + bx + c = 0$
$0 = x^2 - 4x - x + 5 - 1 \implies x^2 - 5x + 4 = 0$
3
$x$
$(x - 1)(x - 4) = 0 \implies x = 1 \text{ or } x = 4$
4
$x$
$\text{For } x = 1: \quad y = (1) + 1 = 2 \\ \text{For } x = 4: \quad y = (4) + 1 = 5$
5
$Finally, state your answer clearly as coordinate pairs. These are the two points where the line and the parabola intersect.$
$\text{The points of intersection are } (1, 2) \text{ and } (4, 5).$

Answer

\text{The points of intersection are } (1, 2) \text{ and } (4, 5).

11. Equations Reducible to Quadratic Form

ax^2+bx+c=0

A visual diagram showing side-by-side comparisons. Left side: . Right side: , with arrows pointing out the parallel structure.

u

Worked example

Worked Example: Solving a Quartic Equation via Substitution

Unmasking the Imposter Equation 🕵️‍♀️

2x^4 - 5x^2 - 12 = 0

1
$x^4$
$\text{Let } u = x^2. \text{ Therefore, } u^2 = (x^2)^2 = x^4.$
2
$u$
$2(x^4) - 5(x^2) - 12 = 0 \implies 2u^2 - 5u - 12 = 0$
3
$u$
$2u^2 - 8u + 3u - 12 = 0 \\ 2u(u-4) + 3(u-4) = 0 \\ (2u+3)(u-4) = 0 \\ \text{So, } u = -\frac{3}{2} \text{ or } u = 4.$
4
$u$
$\text{Case 1: } x^2 = u = -\frac{3}{2} \\ \text{This has no real solutions, as the square of a real number cannot be negative.} \\ \text{Case 2: } x^2 = u = 4$
5
$x$
$x^2 = 4 \implies x = \pm\sqrt{4} \\ x = 2 \text{ or } x = -2$

Answer

x^2 = 4 \implies x = \pm\sqrt{4} \\ x = 2 \text{ or } x = -2

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