More about Equations: Beyond the Straight and Narrow
Hey there! Welcome to the study notes for "More about Equations". In your junior forms, you became an expert at solving linear equations. Now, we're going to level up! This chapter is all about tackling more complex and interesting equations. We'll learn how to find where a straight line crosses a curve, and how to solve tricky equations that are secretly simple quadratics in disguise.
Why is this important? These skills are like a super-tool for problem-solving, not just in maths but in fields like physics, engineering, and economics. Let's get started!
1. When Lines Meet Curves: Solving Simultaneous Equations
Imagine a straight road and a winding, curved river. At what points do they cross? That's exactly what we're finding when we solve simultaneous equations with one linear (the road) and one quadratic (the river) equation.
A Quick Refresher: What are Simultaneous Equations?
Solving simultaneous equations means finding the set of values (like an `x` and a `y`) that make BOTH equations true at the same time. Geometrically, this means finding the coordinates of the point(s) of intersection of their graphs. You've done this with two straight lines, which can only cross at one point (unless they're the same line!). But when a line meets a parabola, things get more interesting.
The Graphical Method (Learning Objective 5.1)
This method is all about visualization. The solutions are simply the points where the graphs of the two equations physically cross on the coordinate plane.
How to do it, step-by-step:
- Carefully draw the graph of the linear equation (the straight line).
- On the same axes, draw the graph of the quadratic equation, which will be a parabola of the form $$y = ax^2 + bx + c$$.
- Look for where the line and the parabola intersect.
- Read the (x, y) coordinates of these intersection points. These are your solutions!
There are three possibilities you can see:
- Two points of intersection: This means there are two distinct real solutions.
- One point of intersection: The line just touches the parabola at one point (it's a tangent). This means there is one repeated real solution.
- No points of intersection: The line and parabola never meet. This means there are no real solutions.
Quick Review Box
Graphical Method: Draw both graphs. The solutions are the coordinates of the points where they cross. Simple as that! In DSE, the parabola will be in the friendly $$y = ax^2 + bx + c$$ form for this method.
The Algebraic Method (Learning Objective 5.2)
This is the more precise, non-graphical way to solve these problems. The most common technique is the substitution method. Think of it like being a detective: you use a clue from one equation to crack the other.
How to do it, step-by-step:
- You'll have a linear equation and a quadratic one. Label them if it helps.
- Take the linear equation and make one variable the subject. For example, rearrange `x - y = 2` into `x = y + 2`. This is your "substitution tool".
- Substitute this expression into the quadratic equation. This will get rid of one variable, leaving you with a normal quadratic equation with just one variable.
- Solve this new quadratic equation to find the value(s) of that variable (e.g., find all the possible `y` values).
- Substitute the value(s) you just found back into the easy linear equation (your "substitution tool" from step 2 is perfect for this) to find the corresponding values of the other variable.
- Write your final answers as coordinate pairs, like `(x, y)`. Make sure you pair them up correctly!
Example: Solve the simultaneous equations:
(1) $$y - x = 1$$
(2) $$y = x^2 - 1$$
Step 1 & 2: From the linear equation (1), let's make `y` the subject.
$$y = x + 1$$
Step 3: Substitute this into the quadratic equation (2).
$$(x + 1) = x^2 - 1$$
Step 4: Rearrange into a standard quadratic and solve.
$$0 = x^2 - x - 2$$
$$0 = (x - 2)(x + 1)$$
So, $$x = 2$$ or $$x = -1$$.
Step 5: Now find the `y` for each `x` using the easy linear rearrangement $$y = x + 1$$.
When $$x = 2$$, $$y = 2 + 1 = 3$$.
When $$x = -1$$, $$y = -1 + 1 = 0$$.
Step 6: State the solutions clearly.
The solutions are (2, 3) and (-1, 0).
Common Mistakes to Avoid:
- Forgetting the second variable: After finding all the `x` values, it's easy to forget to go back and find the corresponding `y` values.
- Pairing them wrong: Make sure the `x` and `y` values that belong together are written as a pair.
- Expansion errors: Be very careful with negative signs and brackets when you substitute and expand.
Key Takeaway: To solve one linear and one quadratic equation, you're finding where a line and a parabola cross. You can do this by drawing them (Graphical Method) or, more accurately, by using algebra (Algebraic Method). The key to the algebraic method is substitution!
2. Master of Disguise: Equations that are Secretly Quadratic
Some equations look really complicated at first glance (with fractions, powers, or logs), but they're often just simple quadratic equations wearing a costume. Our job is to unmask them using a clever trick: substitution.
The "Let u = ..." Trick
If you spot an equation where one term is the square of another term, like $$a(\text{something})^2 + b(\text{something}) + c = 0$$, you can simplify it.
Just say: "Let u = (the 'something')".
Your scary equation will magically transform into the familiar $$au^2 + bu + c = 0$$. Solve for `u`, and then substitute back to find your original variable. Don't worry, this gets much easier with practice!
Types of Disguised Equations (Learning Objective 5.3)
A. Fractional Equations
Example: Solve $$x + \frac{6}{x} = 5$$
- Get rid of the fraction. First, note that `x` cannot be 0. Then, multiply the entire equation by the denominator (`x`).
$$x(x) + x(\frac{6}{x}) = x(5)$$
$$x^2 + 6 = 5x$$ - Rearrange and solve.
$$x^2 - 5x + 6 = 0$$
$$(x - 2)(x - 3) = 0$$
So, $$x = 2$$ or $$x = 3$$. (Both are valid, as neither is 0).
B. Exponential Equations
Example: Solve $$4^x - 3 \cdot 2^x - 4 = 0$$
- Spot the disguise. Notice that $$4^x = (2^2)^x = (2^x)^2$$. So the equation is really:
$$(2^x)^2 - 3(2^x) - 4 = 0$$ - Use the "Let u = ..." trick. Let $$u = 2^x$$. The equation becomes:
$$u^2 - 3u - 4 = 0$$ - Solve for u.
$$(u - 4)(u + 1) = 0$$
So, $$u = 4$$ or $$u = -1$$. - Substitute back and solve for x.
Case 1: $$u = 4 \implies 2^x = 4 \implies 2^x = 2^2 \implies x = 2$$.
Case 2: $$u = -1 \implies 2^x = -1$$. This has no real solution, because a positive number raised to any power can never be negative. - The only solution is $$x = 2$$.
Memory Aid: A positive base like $$a^x$$ is ALWAYS POSITIVE. If you get $$a^x = \text{negative number}$$ or $$a^x = 0$$, reject that solution immediately!
C. Logarithmic Equations
Example: Solve $$\log(x-2) + \log(x+1) = \log 10$$ (Remember, log means log base 10)
- Combine the logs. Use the log law $$\log A + \log B = \log(AB)$$.
$$\log((x-2)(x+1)) = \log 10$$ - Equate the arguments. If $$\log M = \log N$$, then $$M=N$$.
$$(x-2)(x+1) = 10$$ - Expand and solve the quadratic.
$$x^2 - x - 2 = 10$$
$$x^2 - x - 12 = 0$$
$$(x - 4)(x + 3) = 0$$
So, $$x = 4$$ or $$x = -3$$. - CRUCIAL CHECK! The argument of a log must be positive.
Test $$x=4$$: $$\log(4-2) = \log(2)$$ (OK), $$\log(4+1) = \log(5)$$ (OK). So $$x=4$$ is a valid solution.
Test $$x=-3$$: $$\log(-3-2) = \log(-5)$$ (NOT OK!). We must reject this solution. - The only solution is $$x = 4$$.
D. Trigonometric Equations
Example: Solve $$2\sin^2\theta - 5\sin\theta + 2 = 0$$ for $$0^\circ \le \theta \le 360^\circ$$
- Spot the disguise. This looks like a quadratic in terms of $$\sin\theta$$.
Let $$u = \sin\theta$$. The equation becomes:
$$2u^2 - 5u + 2 = 0$$ - Solve for u.
$$(2u - 1)(u - 2) = 0$$
So, $$u = \frac{1}{2}$$ or $$u = 2$$. - Substitute back and solve for θ.
Case 1: $$\sin\theta = \frac{1}{2}$$. The basic angle is $$30^\circ$$. Sine is positive in Quadrants I and II, so $$\theta = 30^\circ$$ and $$\theta = 180^\circ - 30^\circ = 150^\circ$$.
Case 2: $$\sin\theta = 2$$. This has no solution because the sine function can only have values between -1 and 1. - The solutions are $$\theta = 30^\circ, 150^\circ$$.
Key Takeaway: Many complex equations are just quadratics in disguise. The key is to spot the pattern $$a(\text{something})^2 + b(\text{something}) + c = 0$$ and use a substitution (like `let u = ...`) to simplify it. ALWAYS remember to check your final answers to see if they are valid for the original equation (e.g., no division by zero, log of a positive number, sinθ within range).
3. Maths in Action: Solving Real-World Problems (Learning Objective 5.4)
This is where we take all our new algebra skills and apply them to solve practical word problems. The method is the same no matter how complicated the problem seems. You've got this!
A Four-Step Strategy for Word Problems:
- Understand and Define: Read the problem carefully to figure out what you need to find. Then, define your unknowns with variables, like "Let `x` be the width in cm."
- Formulate Equations: Translate the sentences and relationships in the problem into mathematical equations. This is often the trickiest part, so look for keywords like 'sum', 'product', 'perimeter', 'area', etc.
- Solve: Use the algebraic methods you've learned in this chapter to solve the equation or system of equations you created.
- Check and Conclude: Look at your answer. Does it make sense in the real world? (e.g., length can't be negative). Finally, write a clear sentence that answers the original question.
Example: The length of a rectangle is 3 cm more than its width. Its area is 40 cm². Find the dimensions of the rectangle.
1. Understand and Define: We need to find the length and width.
Let the width be `w` cm.
Since the length is 3 cm more than the width, the length is `(w + 3)` cm.
2. Formulate Equation: We are given the area. The formula for the area of a rectangle is Length × Width.
$$(w + 3) \times w = 40$$
3. Solve: This is an equation that can be transformed into a quadratic.
$$w^2 + 3w = 40$$
$$w^2 + 3w - 40 = 0$$
$$(w + 8)(w - 5) = 0$$
So, $$w = -8$$ or $$w = 5$$.
4. Check and Conclude: A width cannot be a negative number, so we reject $$w = -8$$.
The only valid solution is $$w = 5$$.
If the width is 5 cm, the length is $$w + 3 = 5 + 3 = 8$$ cm.
Let's check the area: $$5 \times 8 = 40$$. It works!
Conclusion: The width of the rectangle is 5 cm and the length is 8 cm.
Key Takeaway: Word problems are a process: Understand, Formulate, Solve, and Check. Turning the words into the correct equations is the most important skill to practice. Don't give up!