Strand 4 — Algebra

Equations

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to solve linear equations.
  • By the end of this lesson students will be able to solve quadratic equations using factorisation, the quadratic formula, and completing the square.
  • By the end of this lesson students will be able to solve cubic equations using the Factor Theorem and synthetic division.
  • By the end of this lesson students will be able to solve simultaneous equations, including linear-linear, linear-quadratic, and 3x3 linear systems (Higher Level).
  • By the end of this lesson students will be able to solve equations involving surds and rational expressions, checking for extraneous or invalid solutions.

Key concepts

Linear Equation

An equation of the form ax + b = 0, where a ≠ 0. It has exactly one solution (root).

Quadratic Equation

An equation of the form ax² + bx + c = 0, where a ≠ 0. It can be solved by factorisation, the quadratic formula, or completing the square. It has at most two real roots.

Quadratic Formula

For a quadratic equation in the form ax² + bx + c = 0, the solutions (roots) are given by this formula.

x = (-b ± √(b² - 4ac)) / 2a
Discriminant

The expression b² - 4ac from the quadratic formula. It determines the nature of the roots of a quadratic equation: - If b² - 4ac > 0, there are two distinct real roots. - If b² - 4ac = 0, there is one real root (a repeated root). - If b² - 4ac < 0, there are no real roots (two complex conjugate roots).

Δ = b² - 4ac
Cubic Equation

An equation of the form ax³ + bx² + cx + d = 0, where a ≠ 0. It has up to three real roots.

Factor Theorem

If (x - k) is a factor of a polynomial P(x), then P(k) = 0. Conversely, if P(k) = 0, then (x - k) is a factor of P(x). This is crucial for solving cubic and higher-order polynomial equations.

Simultaneous Equations

A set of equations with multiple variables that must be solved together to find values for all variables that satisfy all equations. Common methods include substitution and elimination. Solving 3x3 linear systems is typically a Higher Level topic.

Equations with Surds

Equations where the variable appears under a square root (or other root) symbol. To solve, isolate the surd and then square both sides. It is essential to check all potential solutions in the original equation for extraneous solutions.

Equations with Rational Expressions

Equations involving algebraic fractions. To solve, find a common denominator and clear the denominators, or multiply by the lowest common multiple (LCM) of the denominators. Always check that potential solutions do not make any original denominator equal to zero.

Extraneous Solutions

Solutions that arise during the algebraic process (e.g., by squaring both sides of an equation or clearing denominators) but do not satisfy the original equation. These solutions must be rejected.

Key facts to remember

  • 1The quadratic formula x = (-b ± √(b² - 4ac)) / 2a solves any quadratic equation ax² + bx + c = 0.
  • 2The discriminant, b² - 4ac, determines the nature of the roots of a quadratic equation (real, repeated, or no real roots).
  • 3The Factor Theorem states that if P(k) = 0, then (x - k) is a factor of the polynomial P(x).
  • 4When solving equations involving surds, always check for extraneous solutions by substituting your answers back into the original equation.
  • 5When solving equations with rational expressions, ensure that no solution makes any original denominator equal to zero.
  • 6Simultaneous linear equations can be solved using substitution or elimination methods.
  • 7Cubic equations have at most three real roots.
  • 8Solving 3x3 simultaneous linear equations is typically a Higher Level topic and often involves systematic elimination.

Worked examples

Example 1

Solve the quadratic equation 3x² - 7x + 2 = 0.

IIdentify the coefficients: a = 3, b = -7, c = 2.
IIApply the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
IIISubstitute the values: x = ( -(-7) ± √((-7)² - 4(3)(2)) ) / (2(3)).
IVSimplify the expression under the square root: x = ( 7 ± √(49 - 24) ) / 6.
VContinue simplifying: x = ( 7 ± √25 ) / 6.
VICalculate the square root: x = ( 7 ± 5 ) / 6.
VIIFind the two possible solutions:
VIIIx₁ = (7 + 5) / 6 = 12 / 6 = 2.
9x₂ = (7 - 5) / 6 = 2 / 6 = 1/3.

Answer

x = 2 or x = 1/3

The quadratic formula is reliable for all quadratic equations, even those that cannot be easily factorised.

Example 2

Solve the cubic equation x³ - 4x² + x + 6 = 0.

ILet P(x) = x³ - 4x² + x + 6.
IIUse the Factor Theorem to test integer factors of the constant term (6): ±1, ±2, ±3, ±6.
IIITest P(1): P(1) = (1)³ - 4(1)² + (1) + 6 = 1 - 4 + 1 + 6 = 4 ≠ 0.
IVTest P(-1): P(-1) = (-1)³ - 4(-1)² + (-1) + 6 = -1 - 4 - 1 + 6 = 0.
VSince P(-1) = 0, (x - (-1)), which is (x + 1), is a factor of P(x).
VIUse synthetic division to divide P(x) by (x + 1):
VII -1 | 1 -4 1 6
VIII | -1 5 -6
9 -----------------
10 1 -5 6 0
11The quotient is x² - 5x + 6.
12Now, solve the quadratic equation x² - 5x + 6 = 0.
13Factorise the quadratic: (x - 2)(x - 3) = 0.
14Set each factor to zero to find the roots:
15x - 2 = 0 ⇒ x = 2.
16x - 3 = 0 ⇒ x = 3.
17Combine all roots found.

Answer

x = -1, 2, 3

Always check a few simple integer factors first when using the Factor Theorem. Synthetic division is an efficient way to find the quadratic factor.

Example 3

Solve the simultaneous equations: 1. y = x - 1 2. x² + y² = 5

ISubstitute equation (1) into equation (2) to eliminate y:
IIx² + (x - 1)² = 5.
IIIExpand the squared term: x² + (x² - 2x + 1) = 5.
IVCombine like terms and rearrange into a standard quadratic equation:
V2x² - 2x + 1 - 5 = 0.
VI2x² - 2x - 4 = 0.
VIIDivide by 2 to simplify: x² - x - 2 = 0.
VIIIFactorise the quadratic equation: (x - 2)(x + 1) = 0.
9Solve for x:
10x - 2 = 0 ⇒ x = 2.
11x + 1 = 0 ⇒ x = -1.
12Substitute these x values back into equation (1) (y = x - 1) to find the corresponding y values:
13If x = 2, then y = 2 - 1 = 1.
14If x = -1, then y = -1 - 1 = -2.
15State the solution pairs.

Answer

The solutions are (x, y) = (2, 1) and (x, y) = (-1, -2).

Always pair the x and y values correctly. Each x-value corresponds to a specific y-value.

Example 4

Solve the equation √(3x + 1) = x - 1.

ISquare both sides of the equation to eliminate the surd:
II(√(3x + 1))² = (x - 1)².
III3x + 1 = x² - 2x + 1.
IVRearrange the equation into a standard quadratic form:
V0 = x² - 2x + 1 - 3x - 1.
VI0 = x² - 5x.
VIIFactorise the quadratic equation: x(x - 5) = 0.
VIIISolve for x:
9x = 0.
10x - 5 = 0 ⇒ x = 5.
11Crucially, check these potential solutions in the original equation to identify any extraneous solutions.
12Check x = 0:
13√(3(0) + 1) = 0 - 1.
14√(1) = -1.
151 = -1. This is false, so x = 0 is an extraneous solution and must be rejected.
16Check x = 5:
17√(3(5) + 1) = 5 - 1.
18√(15 + 1) = 4.
19√(16) = 4.
204 = 4. This is true, so x = 5 is a valid solution.

Answer

x = 5

When squaring both sides of an equation, you may introduce extraneous solutions. Always verify your solutions by substituting them back into the original equation.

Common mistakes

  • Not checking for extraneous solutions when squaring both sides of an equation with surds.
  • Not checking for values that make denominators zero in equations with rational expressions, leading to invalid solutions.
  • Making sign errors when expanding brackets or rearranging terms in equations.
  • Forgetting to find all roots for quadratic or cubic equations, especially when factorising.
  • Incorrectly applying the Factor Theorem or making errors in synthetic division.

Exam tips

  • Always show all steps of your working clearly, as marks are often awarded for method.
  • Check your solutions by substituting them back into the original equation, especially for surd and rational equations.
  • Be meticulous with algebraic manipulation, paying close attention to signs and order of operations.
  • For simultaneous equations, clearly label your equations (e.g., (1), (2)) to make your working easy to follow.

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