Welcome to the World of Quadratics!
Welcome! In this chapter, we are going to dive into Quadratics. You might have seen these "U-shaped" curves before—they appear everywhere, from the path of a basketball thrown into a hoop to the shape of satellite dishes. By the end of these notes, you’ll be able to master the algebra behind these curves, find their highest or lowest points, and solve complex equations with confidence. Don’t worry if this seems tricky at first—we’ll break it down step-by-step!
1. Completing the Square
Completing the square is a clever algebraic "makeover." We take a standard quadratic expression \(ax^2 + bx + c\) and rewrite it in the form \(a(x + p)^2 + q\).
Why do we do this?
The completed square form is like a cheat code for sketching graphs. It tells us exactly where the vertex (the turning point) of the graph is without having to do any extra work. For the form \(y = a(x + p)^2 + q\), the vertex is always at \((-p, q)\).
Step-by-Step Process (for \(x^2 + bx + c\)):
1. Look at the number in front of \(x\) (the \(b\) value).
2. Halve it to get \(p\).
3. Write down \((x + p)^2\).
4. Subtract the square of that number: \(-(p)^2\).
5. Add the original constant \(c\).
Example: Complete the square for \(x^2 + 6x + 5\).
1. Half of \(6\) is \(3\).
2. Write \((x + 3)^2\).
3. Subtract \(3^2\) (which is \(9\)): \((x + 3)^2 - 9\).
4. Add the original \(5\): \((x + 3)^2 - 9 + 5\).
5. Final Answer: \((x + 3)^2 - 4\).
The vertex is at \((-3, -4)\).
Quick Review:
• If \(a\) is positive, the graph is a "smiley face" (minimum point).
• If \(a\) is negative, the graph is a "frown" (maximum point).
Key Takeaway: Completing the square reveals the "turning point" of the quadratic graph.
2. The Discriminant: The "Roots" Fortune Teller
Before you spend time solving a quadratic equation, wouldn't it be nice to know if an answer even exists? That is what the discriminant does! The discriminant is the part of the quadratic formula found under the square root: \(b^2 - 4ac\).
The Three Rules:
1. If \(b^2 - 4ac > 0\): There are two distinct real roots (the graph crosses the x-axis twice).
2. If \(b^2 - 4ac = 0\): There is one repeated root (the graph just touches the x-axis at the vertex).
3. If \(b^2 - 4ac < 0\): There are no real roots (the graph stays completely above or below the x-axis).
Did you know? In exams, if a question says a line is a "tangent" to a curve, it means they touch at exactly one point. This is a huge hint to set your discriminant to zero!
Key Takeaway: Use \(b^2 - 4ac\) to determine the number of solutions without solving the equation.
3. Solving Quadratic Inequalities
Solving \(x^2 - 5x + 6 < 0\) is different from solving an equation. You aren't just looking for two numbers; you are looking for a range of numbers.
How to solve them:
1. Find the Critical Values: Treat the inequality like an equation and solve for \(x\) (factorise or use the formula).
2. Sketch it: Draw a quick "u-shape" and mark your critical values on the x-axis.
3. Identify the region:
• If the inequality is \( < 0 \), you want the part of the graph below the x-axis (usually one continuous interval).
• If the inequality is \( > 0 \), you want the parts above the x-axis (usually two separate intervals).
Common Mistake to Avoid:
Never try to solve an inequality like \(x^2 > 4\) by just square-rooting both sides to get \(x > 2\). You will miss the \(x < -2\) part! Always sketch the graph.
Key Takeaway: A sketch is your best friend when solving inequalities.
4. Simultaneous Equations (Linear and Quadratic)
Sometimes you need to find where a straight line (\(y = mx + c\)) meets a quadratic curve. This is where substitution comes in.
Step-by-Step:
1. Rearrange the linear equation to make either \(x\) or \(y\) the subject (e.g., \(y = ...\)).
2. Substitute this into the quadratic equation.
3. Expand and simplify until you have a standard quadratic equation equal to zero.
4. Solve for the first variable, then plug those values back into the linear equation to find the second variable.
Example: \(y = x + 1\) and \(x^2 + y^2 = 25\).
Substitute \(y\): \(x^2 + (x + 1)^2 = 25\).
Expand: \(x^2 + x^2 + 2x + 1 = 25\).
Simplify: \(2x^2 + 2x - 24 = 0\).
Divide by 2: \(x^2 + x - 12 = 0\).
Factorise: \((x + 4)(x - 3) = 0\).
Solutions: \(x = -4\) (so \(y = -3\)) and \(x = 3\) (so \(y = 4\)).
Key Takeaway: Substitute the simple equation into the complicated one.
5. Equations Reducible to Quadratic Form
Some equations look scary because they have powers like \(x^4\) or square roots \(\sqrt{x}\). However, they are often "hidden quadratics."
The "Substitution Trick":
If you see an equation where one power of \(x\) is exactly double the other, you can use a temporary variable, usually \(u\).
Example 1: \(x^4 - 5x^2 + 4 = 0\)
Let \(u = x^2\). Then \(u^2 = x^4\).
The equation becomes: \(u^2 - 5u + 4 = 0\).
Solve for \(u\): \((u - 4)(u - 1) = 0\), so \(u = 4\) or \(u = 1\).
Don't stop here! Remember \(u = x^2\).
So \(x^2 = 4 \rightarrow x = \pm 2\) and \(x^2 = 1 \rightarrow x = \pm 1\).
Example 2: \(x - 5\sqrt{x} + 6 = 0\)
Let \(u = \sqrt{x}\). Then \(u^2 = x\).
Equation: \(u^2 - 5u + 6 = 0\). Solve for \(u\), then square the results to find \(x\).
Memory Aid:
Think of \(u\) as a "placeholder." It makes the equation look easy so you can solve it, but you must always "give back" the original variable at the end!
Key Takeaway: Look for "double powers" and use a substitution like \(u = x^2\) or \(u = \sqrt{x}\) to simplify the problem.