Quadratic, linear and simultaneous

Grade 12 Mathematics: Section 1.2 - Equations (Quadratic, Linear, Simultaneous including Complex Roots)

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Lesson Objectives

  • Solve linear equations in one variable
  • Solve quadratic equations using factorisation, completing the square, and the quadratic formula
  • Determine the nature of roots using the discriminant
  • Solve quadratic equations with complex (non-real) roots
  • Solve simultaneous linear equations in two variables
  • Solve simultaneous equations involving one linear and one quadratic equation
  • Apply equation-solving techniques to real-world problems

Introduction to Equations

Equations are mathematical statements that assert the equality of two expressions. Solving equations is a fundamental skill in mathematics, allowing us to find unknown values that satisfy given conditions. This section covers linear equations (first degree), quadratic equations (second degree), simultaneous equations (systems of equations), and extends to complex roots when quadratic equations have no real solutions.

Key Formulas for This Section
Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Discriminant: $\Delta = b^2 - 4ac$
$\Delta > 0$: two distinct real roots
$\Delta = 0$: one repeated real root
$\Delta < 0$: two complex conjugate roots
Key Definitions:
• Linear Equation: An equation of the form $ax + b = 0$ where $a \neq 0$.
• Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$ where $a \neq 0$.
• Simultaneous Equations: A set of equations with the same variables that are solved together.
• Complex Root: A root of the form $p + qi$ where $i = \sqrt{-1}$ and $q \neq 0$.
• Discriminant: The expression $\Delta = b^2 - 4ac$ that determines the nature of roots.

Linear Equations

A linear equation in one variable has the general form $ax + b = 0$ where $a \neq 0$. The solution is $x = -\frac{b}{a}$. Linear equations can also involve brackets, fractions, or variables on both sides.

Step-by-Step Method for Solving Linear Equations:
1. Expand any brackets using the distributive law.
2. Collect like terms on each side.
3. Move variable terms to one side and constant terms to the other.
4. Combine like terms.
5. Divide by the coefficient of the variable to isolate it.
Example 1: Simple Linear Equation
Problem: Solve $3x - 7 = 11$.

Solution:
$3x - 7 = 11$
Add 7 to both sides: $3x = 18$
Divide by 3: $x = 6$
Answer: $x = 6$
Example 2: Linear Equation with Brackets
Problem: Solve $4(x - 3) = 2x + 6$.

Solution:
Expand: $4x - 12 = 2x + 6$
Subtract $2x$ from both sides: $2x - 12 = 6$
Add 12: $2x = 18$
Divide by 2: $x = 9$
Answer: $x = 9$
Example 3: Linear Equation with Fractions
Problem: Solve $\frac{x}{2} + \frac{x}{3} = 5$.

Solution:
Multiply by LCM (6): $3x + 2x = 30$
$5x = 30$
$x = 6$
Answer: $x = 6$

Practice for Concept 1 (Linear Equations)

  1. Solve $5x + 3 = 23$.
  2. Solve $2(x - 4) = 3x + 1$.
  3. Solve $\frac{2x}{3} - \frac{x}{2} = 1$.
  4. Solve $4(2x - 1) - 3(x + 2) = 10$.
  5. Solve $7x - 4 = 3x + 12$.

Quadratic Equations

A quadratic equation is of the form $ax^2 + bx + c = 0$ where $a \neq 0$. Solutions can be found by factorisation, completing the square, or using the quadratic formula. The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots.

Methods for Solving Quadratic Equations:
1. Factorisation: If the quadratic factorises nicely, set each factor to zero.
2. Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ — works for all quadratics.
3. Completing the Square: Rewrite as $(x + p)^2 = q$, then solve.
Example 1: Solving by Factorisation
Problem: Solve $x^2 - 5x + 6 = 0$.

Solution:
Factorise: $(x - 2)(x - 3) = 0$
Set each factor to zero: $x - 2 = 0$ or $x - 3 = 0$
$x = 2$ or $x = 3$
Answer: $x = 2, 3$
Example 2: Solving by Quadratic Formula
Problem: Solve $2x^2 - 3x - 2 = 0$.

Solution:
$a = 2$, $b = -3$, $c = -2$
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}$
$x = \frac{3 + 5}{4} = \frac{8}{4} = 2$ or $x = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}$
Answer: $x = 2, -\frac{1}{2}$
Example 3: Solving by Completing the Square
Problem: Solve $x^2 + 6x + 2 = 0$.

Solution:
$x^2 + 6x = -2$
Add $(\frac{6}{2})^2 = 9$ to both sides: $x^2 + 6x + 9 = 7$
$(x + 3)^2 = 7$
$x + 3 = \pm \sqrt{7}$
$x = -3 \pm \sqrt{7}$
Answer: $x = -3 + \sqrt{7}$ or $x = -3 - \sqrt{7}$
Example 4: Using the Discriminant
Problem: Determine the nature of roots for $x^2 - 4x + 4 = 0$.

Solution:
$a = 1$, $b = -4$, $c = 4$
$\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$
Since $\Delta = 0$, there is one repeated real root.
Solve: $(x - 2)^2 = 0$ → $x = 2$ (double root)
Watch Out!
When using the quadratic formula, ensure the equation is in the form $ax^2 + bx + c = 0$ (set equal to zero). Also, be careful with signs when substituting negative values of $b$.

Practice for Concept 2 (Quadratic Equations)

  1. Solve $x^2 - 7x + 10 = 0$ by factorisation.
  2. Solve $3x^2 - 5x - 2 = 0$ using the quadratic formula.
  3. Solve $x^2 + 4x - 3 = 0$ by completing the square.
  4. Find the discriminant of $2x^2 - 4x + 1 = 0$ and state the nature of roots.
  5. Solve $4x^2 - 12x + 9 = 0$.

Quadratic Equations with Complex Roots

When the discriminant $\Delta = b^2 - 4ac$ is negative, the quadratic equation has no real solutions. Instead, it has two complex conjugate roots of the form $p + qi$ and $p - qi$, where $i = \sqrt{-1}$.

Step-by-Step Method for Complex Roots:
1. Identify $a$, $b$, $c$ and compute $\Delta = b^2 - 4ac$.
2. If $\Delta < 0$, write $\sqrt{\Delta} = \sqrt{-\Delta} \cdot i$.
3. Substitute into the quadratic formula: $x = \frac{-b \pm i\sqrt{-\Delta}}{2a}$.
4. Simplify to the form $p \pm qi$.
Example 1: Quadratic with Complex Roots
Problem: Solve $x^2 + 4x + 5 = 0$.

Solution:
$a = 1$, $b = 4$, $c = 5$
$\Delta = 16 - 4(1)(5) = 16 - 20 = -4$
$\sqrt{\Delta} = \sqrt{-4} = 2i$
$x = \frac{-4 \pm 2i}{2} = -2 \pm i$
Answer: $x = -2 + i$ or $x = -2 - i$
Example 2: Complex Roots with a > 1
Problem: Solve $2x^2 - 3x + 4 = 0$.

Solution:
$a = 2$, $b = -3$, $c = 4$
$\Delta = (-3)^2 - 4(2)(4) = 9 - 32 = -23$
$\sqrt{\Delta} = i\sqrt{23}$
$x = \frac{3 \pm i\sqrt{23}}{4}$
Answer: $x = \frac{3}{4} \pm \frac{\sqrt{23}}{4}i$
Example 3: Perfect Square Complex Roots
Problem: Solve $x^2 - 2x + 2 = 0$.

Solution:
$a = 1$, $b = -2$, $c = 2$
$\Delta = 4 - 8 = -4$
$\sqrt{\Delta} = 2i$
$x = \frac{2 \pm 2i}{2} = 1 \pm i$
Answer: $x = 1 + i$ or $x = 1 - i$
Example 4: Sum and Product of Complex Roots
Problem: For $x^2 + 2x + 5 = 0$, find the sum and product of roots.

Solution:
Sum of roots $= -\frac{b}{a} = -\frac{2}{1} = -2$
Product of roots $= \frac{c}{a} = \frac{5}{1} = 5$
The roots are $-1 \pm 2i$. Check: $(-1+2i)+(-1-2i) = -2$, $(-1+2i)(-1-2i) = 1 + 4 = 5$ ✓
Watch Out!
Complex roots always occur in conjugate pairs when coefficients are real. Remember that $i^2 = -1$ and be careful with arithmetic involving $i$. Simplify complex roots to the form $p + qi$ where $p$ and $q$ are real numbers.
Nature of Roots Summary
Discriminant ($\Delta$) Nature of Roots Example
$\Delta > 0$ Two distinct real roots $x^2 - 3x + 2 = 0$ → $x = 1, 2$
$\Delta = 0$ One repeated real root $x^2 - 4x + 4 = 0$ → $x = 2$ (double)
$\Delta < 0$ Two complex conjugate roots $x^2 + 2x + 5 = 0$ → $x = -1 \pm 2i$

Practice for Concept 3 (Complex Roots)

  1. Solve $x^2 + 2x + 2 = 0$.
  2. Solve $x^2 - 4x + 8 = 0$.
  3. Solve $2x^2 - 2x + 1 = 0$.
  4. Solve $x^2 + 9 = 0$.
  5. Find the complex roots of $3x^2 - x + 2 = 0$.

Core Concept 4: Simultaneous Equations

Simultaneous equations involve two or more equations with the same variables. Solutions satisfy all equations simultaneously. Methods include substitution, elimination, and graphical methods.

Methods for Solving Simultaneous Equations:
1. Substitution Method: Solve one equation for one variable, substitute into the other.
2. Elimination Method: Add or subtract equations to eliminate one variable.
3. For Linear + Quadratic: Substitute the linear equation into the quadratic equation.
Example 1: Linear Simultaneous Equations (Elimination)
Problem: Solve $2x + y = 7$ and $x - y = 2$.

Solution:
Add the equations: $(2x + y) + (x - y) = 7 + 2$ → $3x = 9$ → $x = 3$
Substitute $x = 3$ into $x - y = 2$: $3 - y = 2$ → $y = 1$
Answer: $x = 3$, $y = 1$
Example 2: Linear Simultaneous Equations (Substitution)
Problem: Solve $y = 2x + 1$ and $3x + 2y = 9$.

Solution:
Substitute $y = 2x + 1$ into $3x + 2y = 9$:
$3x + 2(2x + 1) = 9$ → $3x + 4x + 2 = 9$ → $7x = 7$ → $x = 1$
Then $y = 2(1) + 1 = 3$
Answer: $x = 1$, $y = 3$
Example 3: One Linear, One Quadratic
Problem: Solve $y = x + 1$ and $y = x^2 - 3x + 4$.

Solution:
Substitute: $x + 1 = x^2 - 3x + 4$
Rearrange: $0 = x^2 - 3x + 4 - x - 1 = x^2 - 4x + 3$
Factorise: $(x - 1)(x - 3) = 0$ → $x = 1$ or $x = 3$
When $x = 1$, $y = 1 + 1 = 2$
When $x = 3$, $y = 3 + 1 = 4$
Answer: $(1, 2)$ and $(3, 4)$
Example 4: Linear + Quadratic with No Real Solution
Problem: Solve $y = 2x + 1$ and $y = x^2 + 1$.

Solution:
Substitute: $2x + 1 = x^2 + 1$ → $2x = x^2$ → $x^2 - 2x = 0$ → $x(x - 2) = 0$
$x = 0$ or $x = 2$
When $x = 0$, $y = 1$; when $x = 2$, $y = 5$
Answer: $(0, 1)$ and $(2, 5)$ — two intersection points.
Example 5: System with No Real Intersection
Problem: Solve $y = x + 2$ and $y = x^2 + 1$.

Solution:
$x + 2 = x^2 + 1$ → $0 = x^2 - x - 1$
$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$ (real roots)
So there are two real intersection points. If discriminant were negative, there would be no real solutions.
Watch Out!
When solving simultaneous equations with a quadratic, you may get extraneous solutions if you square both sides. Always check your solutions in the original equations. For linear-quadratic systems, there can be 0, 1, or 2 real solutions.

Practice for Concept 4 (Simultaneous Equations)

  1. Solve: $3x + y = 10$ and $2x - y = 5$.
  2. Solve: $y = 3x - 2$ and $2x + 3y = 16$.
  3. Solve: $y = 2x + 1$ and $y = x^2 - 2x + 3$.
  4. Solve: $x + y = 5$ and $x^2 + y^2 = 13$.
  5. Solve: $y = x^2 - 4$ and $y = 2x - 5$.

Methods & Techniques

Mastering equation solving requires systematic approaches and verification strategies. Use these techniques to ensure accuracy.

Verification / Checking Strategy:
1. For linear equations: Substitute solution back into the original equation.
2. For quadratic equations: Use the sum and product of roots to verify, or substitute back.
3. For complex roots: Check that the sum and product match $-\frac{b}{a}$ and $\frac{c}{a}$.
4. For simultaneous equations: Substitute solutions into both original equations.
Example: Checking Quadratic Solutions
Original problem: Solve $x^2 - 5x + 6 = 0$.
Your solution: $x = 2, 3$

Check:
For $x = 2$: $4 - 10 + 6 = 0$ ✓
For $x = 3$: $9 - 15 + 6 = 0$ ✓
Sum of roots = $5 = -\frac{b}{a} = 5$ ✓
Product = $6 = \frac{c}{a} = 6$ ✓
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Forgetting to set quadratic equations to zero before solving → Solution: Always rearrange to $ax^2 + bx + c = 0$ first.
• Pitfall 2: Sign errors in the quadratic formula with negative $b$ → Solution: Write $-b$ carefully; use parentheses.
• Pitfall 3: Misidentifying complex roots as real → Solution: Check discriminant before solving; expect complex when $\Delta < 0$.
• Pitfall 4: Forgetting to check for extraneous solutions in systems → Solution: Always substitute back into original equations.
• Pitfall 5: Errors with $i = \sqrt{-1}$ simplification → Solution: Remember $i^2 = -1$ and simplify fractions carefully.

Technique Practice

  1. Verify that $x = 3$ satisfies $2x - 5 = 1$.
  2. Check the roots of $x^2 + 4x + 5 = 0$ using sum and product.
  3. Identify the error: For $x^2 + 2x + 2 = 0$, a student wrote $x = -1 \pm 2$. Correct it.
  4. For the simultaneous equations $y = x^2$ and $y = x + 2$, a student found $x = 2, -1$. Check both solutions.

Real-World Applications

Equations are used extensively in physics, engineering, economics, and many other fields to model real-world situations.

Application 1: Projectile Motion (Quadratic)
Scenario: A ball is thrown upward with initial velocity 20 m/s from a height of 1.5 m. Height $h$ in metres after $t$ seconds is $h = -5t^2 + 20t + 1.5$. When does the ball hit the ground?
Problem: Solve $-5t^2 + 20t + 1.5 = 0$.

Solution:
Multiply by -2 or use formula: $5t^2 - 20t - 1.5 = 0$
$a = 5$, $b = -20$, $c = -1.5$
$t = \frac{20 \pm \sqrt{400 - 4(5)(-1.5)}}{10} = \frac{20 \pm \sqrt{400 + 30}}{10} = \frac{20 \pm \sqrt{430}}{10}$
$t \approx \frac{20 \pm 20.736}{10}$ → positive root: $t \approx 4.074$ seconds
Answer: Approximately 4.07 seconds
Application 2: Break-Even Analysis (Linear Simultaneous)
Scenario: A company's cost is $C = 500 + 20x$ and revenue is $R = 50x$, where $x$ is units sold. Find break-even point.
Problem: Solve $500 + 20x = 50x$.

Solution:
$500 = 30x$ → $x = \frac{500}{30} = 16.\dot{6}$ units
Need to sell about 17 units to break even.
Application 3: AC Circuit Analysis (Complex Roots)
Scenario: In an RLC circuit, the impedance equation leads to $x^2 + 2x + 5 = 0$ for the frequency variable. Find the complex frequency.
Problem: Solve $x^2 + 2x + 5 = 0$.

Solution:
$x = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$
The complex roots represent damped oscillatory behaviour in the circuit.
Application 4: Geometry - Rectangle Area
Scenario: A rectangle has length 3 cm more than twice its width. Its area is 65 cm². Find dimensions.
Problem: Let width = $w$, length = $2w + 3$. Area: $w(2w + 3) = 65$.

Solution:
$2w^2 + 3w - 65 = 0$
$w = \frac{-3 \pm \sqrt{9 + 520}}{4} = \frac{-3 \pm \sqrt{529}}{4} = \frac{-3 \pm 23}{4}$
Positive root: $w = \frac{20}{4} = 5$ cm, length $= 2(5) + 3 = 13$ cm
Answer: Width = 5 cm, Length = 13 cm

Cross-Curricular Connections

  • Physics: Kinematics equations ($s = ut + \frac{1}{2}at^2$), Ohm's law, circuit analysis
  • Engineering: Stress-strain relationships, control systems (complex roots indicate oscillations)
  • Economics: Supply-demand equilibrium, profit maximisation
  • Biology: Population growth models, enzyme kinetics

Cumulative Practice Exercises

Try these problems on your own. Show all working steps. Use the verification strategies to check your answers.

  1. Solve $3x - 7 = 2x + 5$.
  2. Solve $x^2 - 6x + 8 = 0$ by factorisation.
  3. Solve $2x^2 - 4x - 3 = 0$ using the quadratic formula.
  4. Determine the nature of roots for $x^2 - 3x + 5 = 0$.
  5. Solve $x^2 + 6x + 10 = 0$ (complex roots).
  6. Solve $3x^2 - 2x + 1 = 0$ (complex roots).
  7. Solve simultaneously: $2x + y = 8$ and $3x - 2y = 5$.
  8. Solve simultaneously: $y = 2x - 1$ and $y = x^2 - 2x + 3$.
  9. Solve simultaneously: $x + y = 7$ and $xy = 10$.
  10. For $x^2 + kx + 9 = 0$, find $k$ so that the roots are equal.
  11. For $x^2 + 4x + k = 0$, find $k$ so that the roots are complex.
  12. The quadratic $x^2 + bx + c = 0$ has roots $2 + 3i$ and $2 - 3i$. Find $b$ and $c$.
  13. A rectangle's length is 5 cm more than its width. The area is 84 cm². Find dimensions.
  14. Error analysis: A student solved $x^2 + 4x + 5 = 0$ and got $x = -2 \pm 1$. Is this correct? Explain.
  15. The sum of two numbers is 12 and the sum of their squares is 80. Find the numbers.
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Answers to Cumulative Exercises

  1. Problem: $3x - 7 = 2x + 5$
    Answer: $x = 12$
  2. Problem: $x^2 - 6x + 8 = 0$
    Answer: $x = 2, 4$
  3. Problem: $2x^2 - 4x - 3 = 0$
    Answer: $x = \frac{4 \pm \sqrt{16 + 24}}{4} = \frac{4 \pm \sqrt{40}}{4} = \frac{4 \pm 2\sqrt{10}}{4} = 1 \pm \frac{\sqrt{10}}{2}$
  4. Problem: $x^2 - 3x + 5 = 0$ discriminant
    Answer: $\Delta = 9 - 20 = -11 < 0$ → complex conjugate roots
  5. Problem: $x^2 + 6x + 10 = 0$
    Answer: $x = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2} = -3 \pm i$
  6. Problem: $3x^2 - 2x + 1 = 0$
    Answer: $\Delta = 4 - 12 = -8$, $x = \frac{2 \pm 2i\sqrt{2}}{6} = \frac{1 \pm i\sqrt{2}}{3}$
  7. Problem: $2x + y = 8$, $3x - 2y = 5$
    Answer: Multiply first by 2: $4x + 2y = 16$; add to second: $7x = 21$ → $x = 3$, $y = 2$
  8. Problem: $y = 2x - 1$, $y = x^2 - 2x + 3$
    Answer: $2x - 1 = x^2 - 2x + 3$ → $0 = x^2 - 4x + 4 = (x - 2)^2$ → $x = 2$, $y = 3$ (one intersection)
  9. Problem: $x + y = 7$, $xy = 10$
    Answer: From $y = 7 - x$, $x(7 - x) = 10$ → $7x - x^2 = 10$ → $x^2 - 7x + 10 = 0$ → $(x - 2)(x - 5) = 0$ → $(2, 5)$ and $(5, 2)$
  10. Problem: $x^2 + kx + 9 = 0$ equal roots → $\Delta = 0$
    Answer: $k^2 - 36 = 0$ → $k = \pm 6$
  11. Problem: $x^2 + 4x + k = 0$ complex roots → $\Delta < 0$
    Answer: $16 - 4k < 0$ → $4k > 16$ → $k > 4$
  12. Problem: Roots $2 \pm 3i$ → sum = $4 = -b$, product = $4 + 9 = 13 = c$
    Answer: $b = -4$, $c = 13$ → $x^2 - 4x + 13 = 0$
  13. Problem: Length = width + 5, area = 84 → $w(w + 5) = 84$ → $w^2 + 5w - 84 = 0$ → $(w + 12)(w - 7) = 0$ → $w = 7$ cm, length = 12 cm
  14. Problem: Error analysis: $x^2 + 4x + 5 = 0$ → $x = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$
    Answer: Student forgot the $i$; correct is $-2 \pm i$
  15. Problem: $x + y = 12$, $x^2 + y^2 = 80$
    Answer: $y = 12 - x$, $x^2 + (12 - x)^2 = 80$ → $x^2 + 144 - 24x + x^2 = 80$ → $2x^2 - 24x + 64 = 0$ → $x^2 - 12x + 32 = 0$ → $(x - 4)(x - 8) = 0$ → numbers are 4 and 8

Conclusion & Summary

Equations are the language of algebra. Linear equations provide the foundation, quadratic equations introduce non-linear relationships, and simultaneous equations allow us to solve for multiple unknowns. Complex roots extend our number system to solve equations that have no real solutions.

Key Takeaways:
1. Linear equations: Solve by isolating the variable using inverse operations.
2. Quadratic equations: Use factorisation, quadratic formula, or completing the square.
3. Discriminant $\Delta$: $\Delta > 0$ → two real roots; $\Delta = 0$ → one repeated root; $\Delta < 0$ → complex conjugate roots.
4. Complex roots: Occur in conjugate pairs $p \pm qi$ when coefficients are real.
5. Simultaneous equations: Use substitution or elimination; linear-quadratic systems may have 0, 1, or 2 solutions.
6. Verification: Always check solutions in original equations.

Keep practicing with different types of equations. Mastery of equation solving is essential for calculus, physics, engineering, and beyond!

Video Resource

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