🎓 Pure Mathematics 1: Chapter 1.1 Quadratics Study Notes 📚

Hello and welcome to the world of Quadratics! This chapter is absolutely fundamental to your AS/A Level journey. Quadratics describe a specific type of curve called a parabola, which you see everywhere—from the path of a thrown ball to the shape of satellite dishes. Mastering these skills will not only secure marks in Paper 1 but also provide essential tools for later topics like Coordinate Geometry and Differentiation.

Let's dive in and tackle these powerful equations step-by-step!

1. Understanding the Quadratic Function and its Graph

Standard Form

A quadratic function is a polynomial of degree 2, always written in the standard form:

\[y = ax^2 + bx + c\]

where \(a\), \(b\), and \(c\) are constants, and crucially, \(a \neq 0\).

The Parabola (The Shape of the Graph)

  • The graph of a quadratic is called a parabola.
  • It has a single turning point called the vertex.

Determining the Shape using a:

  • If \(a > 0\) (a is positive), the parabola opens upwards. It has a U-shape, or a "happy face" 😊. The vertex is a minimum point.
  • If \(a < 0\) (a is negative), the parabola opens downwards. It has an inverted U-shape, or a "sad face" 😞. The vertex is a maximum point.

Key Takeaway: The standard form \(ax^2 + bx + c\) tells you the coefficients you need for the formula, but the sign of \(a\) immediately tells you the shape of the graph (minimum or maximum).

2. Completing the Square (C.T.S.)

Completing the square is an algebraic technique used for two main purposes in quadratics:

  1. To easily identify the vertex (turning point) of the graph.
  2. To solve quadratic equations when factorizing is difficult or impossible.

The Completed Square Form

Any quadratic \(ax^2 + bx + c\) can be rewritten in the form:

\[a(x+p)^2 + q\]

The vertex of the parabola is located at the point \((-p, q)\).

Step-by-Step: Completing the Square (when a = 1)

Let's use the example \(y = x^2 + 6x + 5\):

  1. Half the \(b\) term: Take the coefficient of \(x\) (\(b\)) and divide it by 2. (Here, \(6/2 = 3\)). This is your \(p\).
  2. Write the bracket: Start by writing the squared term \((x+3)^2\).
  3. Subtract the square: \((x+3)^2\) expands to \(x^2 + 6x + 9\). You need to balance the equation by subtracting the extra constant you introduced (\(9\)).
  4. Add the original \(c\) term: Add the original constant from the question (\(+5\)).
  5. Simplify:

    \(y = (x+3)^2 - 9 + 5\)
    \(y = (x+3)^2 - 4\)

Did you know? Since \(p=3\) and \(q=-4\), the vertex is at \((-3, -4)\).

Handling \(a \neq 1\)

If \(a\) is not 1 (e.g., \(2x^2 - 8x + 3\)), first factor out \(a\) from the \(x^2\) and \(x\) terms only:

\(2x^2 - 8x + 3 = 2(x^2 - 4x) + 3\)

Now, complete the square inside the bracket for \((x^2 - 4x)\):

\(2\left[(x - 2)^2 - 4\right] + 3\)

Distribute the 2 back in:

\(2(x - 2)^2 - 8 + 3 = 2(x - 2)^2 - 5\)

Vertex is at \((2, -5)\).

Key Takeaway: C.T.S. puts the quadratic into vertex form. Remember the trick: half the middle term, square it, and subtract it to keep the expression balanced!

3. The Discriminant (\(b^2 - 4ac\))

The discriminant, denoted by the Greek letter delta (\(\Delta\)), is the key part hidden under the square root in the quadratic formula. It tells you the nature of the roots (how many times the graph crosses the x-axis) without having to solve the equation.

\[\Delta = b^2 - 4ac\]

The Three Cases:

  1. \(\Delta > 0\) (Positive Discriminant):
    • Meaning: Two distinct real roots.
    • Graph: The parabola crosses the x-axis at two separate points.
  2. \(\Delta = 0\) (Zero Discriminant):
    • Meaning: One real root (a repeated root, or equal roots).
    • Graph: The parabola touches the x-axis exactly once (it is tangential to the x-axis) at its vertex.
  3. \(\Delta < 0\) (Negative Discriminant):
    • Meaning: No real roots (the roots are complex, but for P1, we just say "no real roots").
    • Graph: The parabola lies entirely above the x-axis (if \(a>0\)) or entirely below the x-axis (if \(a<0\)).
Application: Lines and Curves

This is where the discriminant becomes incredibly useful! If you are asked to determine how many times a line \(y = mx + c\) intersects a quadratic curve \(y = ax^2 + bx + d\), you set the equations equal to each other:

\[ax^2 + bx + d = mx + c\]

Rearrange this into a new quadratic equation in the form \(Ax^2 + Bx + C = 0\). Then, use the discriminant on this new equation to find the number of intersection points.

Example: To find the values of \(k\) for which the line \(y = x + k\) intersects the curve \(y = x^2 - 3x + 1\):

  1. Set them equal: \(x^2 - 3x + 1 = x + k\)
  2. Rearrange: \(x^2 - 4x + (1 - k) = 0\)
  3. For intersection, we need \(\Delta \geq 0\). Use \(A=1\), \(B=-4\), \(C=(1-k)\).
  4. \(\Delta = (-4)^2 - 4(1)(1-k) \geq 0\).

Key Takeaway: The discriminant is your tool for existence and tangency. Remember: Positive means two, zero means one (repeated), negative means none.

4. Solving Quadratic Equations and Inequalities

Solving Equations (\(ax^2 + bx + c = 0\))

There are three ways to solve a quadratic equation, and you must be proficient in all of them:

  1. Factorizing: Fastest method if possible. (E.g., \(x^2 - 5x + 6 = 0\) leads to \((x-2)(x-3) = 0\)).
  2. Completing the Square: Essential if the question asks for the answer in surd form or for the structure of the vertex.
  3. The Quadratic Formula: Works for every quadratic, guaranteed. Essential for cases that don't factorize easily.
The Quadratic Formula (Memorise this!)

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

Don't worry if you forget it—it is provided in the MF19 Formula Booklet! But knowing it saves time.

Common Mistake: Forgetting to divide the entire expression (\(-b\) and the square root) by \(2a\).

Solving Quadratic Inequalities

Solving inequalities (involving \(\leq, <, >, \geq\)) requires an extra step: drawing a sketch!

Step-by-Step for \(x^2 - 5x + 6 > 0\):

  1. Find the roots: Solve the equation \(x^2 - 5x + 6 = 0\). We found the roots are \(x=2\) and \(x=3\).
  2. Sketch the graph: Since \(a=1\) (positive), it's a happy face parabola opening upwards, crossing the axis at 2 and 3.
  3. Identify the required region: The inequality is \(> 0\), meaning we want the values of \(x\) where the graph is above the x-axis.
  4. Write the solution: The graph is above the axis when \(x\) is less than 2 OR when \(x\) is greater than 3.

    Solution: \(x < 2\) or \(x > 3\)

If the inequality was \(x^2 - 5x + 6 \leq 0\), you would be looking for the region below the x-axis, giving the answer \(2 \leq x \leq 3\).

Key Takeaway: Always find the roots first. For inequalities, always sketch the graph. This prevents common sign errors.

5. Simultaneous Equations

You must be able to solve pairs of simultaneous equations where one equation is linear and the other is quadratic.

Method: Substitution

The core idea is to use the linear equation to get an expression for one variable (e.g., \(y\)) and substitute it into the quadratic equation. This reduces the problem to solving a single quadratic equation.

Example:
Equation 1 (Linear): \(x + y = 5\)
Equation 2 (Quadratic): \(x^2 + 2y^2 = 17\)

Step-by-Step Procedure:

  1. Isolate a variable in the linear equation: From (1), we get \(y = 5 - x\).
  2. Substitute into the quadratic equation: Replace \(y\) in (2) with \((5 - x)\):

    \(x^2 + 2(5 - x)^2 = 17\)

  3. Expand and Simplify: Carefully expand \((5 - x)^2 = 25 - 10x + x^2\).

    \(x^2 + 2(25 - 10x + x^2) = 17\)
    \(x^2 + 50 - 20x + 2x^2 = 17\)
    \(3x^2 - 20x + 33 = 0\)

  4. Solve the resulting quadratic: Solve for \(x\) (by factorizing or using the formula). Let's say you find two solutions for \(x\), e.g., \(x_1 = 3\) and \(x_2 = 3.66\).
  5. Find the corresponding pairs for the other variable: Use the simplest equation (Equation 1: \(y = 5 - x\)) to find the corresponding \(y\) values for each \(x\) value.

    If \(x=3\), then \(y = 5 - 3 = 2\). Solution: \((3, 2)\).
    If \(x=3.66\), then \(y = 5 - 3.66 = 1.34\). Solution: \((3.66, 1.34)\).

Accessibility Tip: Always remember the final step! Simultaneous equations usually yield two pairs of solutions, not just two values for one variable. You must present the coordinates clearly.

Key Takeaway: Simultaneous equations always rely on substitution. Don't forget to find the second variable after solving the quadratic!

6. Equations Reducible to Quadratic Form

Some equations might look scary because they have high powers, fractional powers, or trigonometric terms, but they can be simplified into a quadratic equation by using a simple substitution. These are equations that are quadratic in some function of \(x\).

Recognising the Pattern

Look for equations where the power of one term is double the power of another term, like in these examples from the syllabus:

  • \(x^4 - 5x^2 + 4 = 0\) (Here, 4 is double 2.)
  • \(6x + \sqrt{x} - 1 = 0\) (If we let \(u = \sqrt{x}\), then \(u^2 = x\).)
  • \(\tan^2 x = 1 + \tan x\) (Here, the power 2 is double the power 1.)

Solving Reducible Equations (Step-by-Step)

Let's solve \(x^4 - 5x^2 + 4 = 0\):

  1. Identify the base variable: Let \(u = x^2\).
  2. Perform the substitution: Since \(x^4 = (x^2)^2 = u^2\), the equation becomes:

    \[u^2 - 5u + 4 = 0\]

  3. Solve the quadratic for \(u\): Factorizing gives \((u - 4)(u - 1) = 0\).

    Solutions for \(u\): \(u = 4\) or \(u = 1\).

  4. Solve back for \(x\) (The crucial step!): Remember the definition \(u = x^2\). We must substitute back to find the values of \(x\).
    • If \(u = 4\): \(x^2 = 4 \implies x = \pm 2\)
    • If \(u = 1\): \(x^2 = 1 \implies x = \pm 1\)

The original equation has four solutions: \(x = -2, -1, 1, 2\).

Analogy:

Think of substitution as temporary clothes. You change into \(u\) to do the complicated algebra, but you must put your real clothes (\(x\)) back on at the end to answer the question!

Important Consideration (for equations involving roots or restricted domains):

If you had \(6x + \sqrt{x} - 1 = 0\), letting \(u = \sqrt{x}\) means that \(u\) must be positive (\(u \geq 0\)). If your quadratic solution for \(u\) gives a negative value, you must discard that root before solving for \(x\).

Key Takeaway: Use substitution (e.g., \(u = f(x)\)) to turn the hard problem into a simple quadratic, but always check if the new variable \(u\) has any domain restrictions, and always reverse the substitution to find the final answers for \(x\).