Solving Quadratic Equations Involving Fractions Revision Notes for Leaving Cert Mathematics

Solving Quadratic Equations Involving Fractions

What are quadratic equations involving fractions?

Quadratic equations involving fractions are equations that contain algebraic fractions (where the variable appears in the denominator) and lead to quadratic equations when solved. These are also called rational equations.

When we encounter an equation with fractions containing variables in the denominators, we need a systematic approach to solve them effectively.

The LCM method

The key technique for solving these equations is to multiply every term by the LCM (Least Common Multiple) of all denominators. This eliminates the fractions and transforms the equation into a standard quadratic equation.

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The LCM method is the most efficient approach because it removes all fractions in one step, converting the complex rational equation into a familiar quadratic equation that can be solved using standard techniques.

Step-by-step process:

Identify all denominators in the equation
Find the LCM of these denominators
Multiply every term by this LCM
Simplify the resulting equation
Rearrange into standard quadratic form
Solve by factorising or using the quadratic formula

Worked example 1: Simple rational equation

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Worked Example: Simple Rational Equation

Problem: Solve $\frac{x-3}{3} + \frac{12}{x} = 4$

Solution:

Step 1: Identify the denominators: 3 and $x$

Step 2: Find the LCM: $3x$

Step 3: Multiply every term by $3x$ : $\frac{3x(x-3)}{3} + \frac{3x(12)}{x} = 4(3x)$

Step 4: Simplify: $x(x-3) + 3(12) = 12x$ $x^2 - 3x + 36 = 12x$

Step 5: Rearrange to standard form: $x^2 - 3x + 36 - 12x = 0$ $x^2 - 15x + 36 = 0$

Step 6: Factorise and solve: $(x - 3)(x - 12) = 0$

Therefore: $x = 3$ or $x = 12$

Worked example 2: Complex rational equation

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Worked Example: Complex Rational Equation

Problem: Solve $\frac{2}{x-1} - \frac{1}{x+2} = \frac{1}{2}$

Solution:

Step 1: Identify the denominators: $(x-1)$ , $(x+2)$ , and $2$

Step 2: Find the LCM: $2(x-1)(x+2)$

Step 3: Multiply every term by $2(x-1)(x+2)$ : $\frac{2(x-1)(x+2) \cdot 2}{x-1} - \frac{2(x-1)(x+2) \cdot 1}{x+2} = \frac{2(x-1)(x+2) \cdot 1}{2}$

Step 4: Simplify: $4(x+2) - 2(x-1) = (x-1)(x+2)$ $4x + 8 - 2x + 2 = x^2 + 2x - x - 2$ $2x + 10 = x^2 + x - 2$

Step 5: Rearrange to standard form: $0 = x^2 + x - 2 - 2x - 10$ $0 = x^2 - x - 12$

Step 6: Factorise and solve: $(x - 4)(x + 3) = 0$

Therefore: $x = 4$ or $x = -3$

Key exam tips

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Essential Exam Tips:

Always check your solutions by substituting back into the original equation
Be careful with signs when multiplying through by the LCM
Watch out for denominators that equal zero - these values cannot be solutions
Remember to rearrange into standard quadratic form before factorising
Show all working clearly - marks are awarded for method as well as the final answer

Common mistakes to avoid

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Common Pitfalls to Avoid:

Forgetting to multiply ALL terms by the LCM
Making algebraic errors when expanding brackets
Not checking that solutions don't make denominators zero
Rushing the factorisation step

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Key Points to Remember:

Find the LCM of all denominators first
Multiply every single term by this LCM to clear fractions
The result will always be a quadratic equation that can be solved by factorising
Always verify your solutions by substituting back into the original equation
Check that solutions don't make any denominator equal to zero

Solving Quadratic Equations Involving Fractions (Leaving Cert Mathematics): Revision Notes