Quadratic Equations with fractions (Junior Cert Mathematics): Revision Notes

Example Problem:

Solve: $\frac{3}{x-1} + \frac{2}{x+2} = 1$

In this example, we have a quadratic equation involving fractions. Our task is to eliminate the fractions first, simplify the equation, and then solve it.

Step 1: Eliminate the Fractions

What we do:

To eliminate the fractions, we first find the common denominator for all the fractions involved. The common denominator here would be $(x-1)(x+2)$.

Why we do it:

Multiplying every term in the equation by this common denominator will remove all the fractions, making the equation easier to solve.

Step 2: Multiply Every Term by the Common Denominator

What we do:

Multiply each term in the equation by the common denominator $(x-1)(x+2)$ to eliminate the fractions.

$\left(\frac{3}{x-1} \right) \times (x-1)(x+2) + \left(\frac{2}{x+2} \right) \times (x-1)(x+2) = 1 \times (x-1)(x+2)$

Why we do it:

This step clears out the denominators, allowing us to focus on the numerators. After multiplying, the equation becomes:

$3(x+2) + 2(x-1) = (x-1)(x+2)$

Step 3: Expand and Simplify the Equation

What we do:

Expand both sides of the equation to remove the brackets and simplify the equation.

Expanding the left side: $3x + 6 + 2x - 2 = x^2 + 2x - x - 2$

Simplifying both sides: $5x + 4 = x^2 + x - 2$

Step 4: Rearrange into Standard Quadratic Form

What we do:

Move all terms to one side of the equation to form a standard quadratic equation.

Subtract $5x + 4$ from both sides: $0 = x^2 - 4x - 6$

Rewriting it: $x^2 - 4x - 6 = 0$

Now, the equation is in standard quadratic form.

Step 5: Solve Using the $-b$ Formula

What we do:

Now that we have the equation in standard form, we can use the $-b$ formula to solve it.

First, identify $a$, $b$, and $c$:

• $a = 1$
• $b = -4$
• $c = -6$ Substitute these into the formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-6)}}{2(1)}$ Simplifying: $x = \frac{4 \pm \sqrt{16 + 24}}{2}$ $x = \frac{4 \pm \sqrt{40}}{2}$

Since $\sqrt{40}$ simplifies to $2\sqrt{10}$, the equation becomes: $x = \frac{4 \pm 2\sqrt{10}}{2}$ $x = 2 \pm \sqrt{10}$

So, the solutions are: $x = 2 + \sqrt{10} \quad \text{or} \quad x = 2 - \sqrt{10}$

These are real solutions, and you can simplify further if needed, but they are already in their exact form.