Using the Quadratic Formula Revision Notes for Leaving Cert Mathematics

Using the Quadratic Formula

What is the quadratic formula?

The quadratic formula is a mathematical tool used to solve quadratic equations when factorisation is not possible or practical. A quadratic equation has the standard form $ax^2 + bx + c = 0$ , where a, b, and c are coefficients and a ≠ 0.

When the expression $ax^2 + bx + c$ cannot be factorised easily, we can use the quadratic formula to find the exact solutions (roots) of the equation.

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The quadratic formula is particularly useful when dealing with equations that don't factor neatly or when you need precise decimal answers. While factorisation might be quicker for simple cases, the quadratic formula provides a systematic approach that works for any quadratic equation.

The quadratic formula

The quadratic formula states that for any quadratic equation in the form $ax^2 + bx + c = 0$ , the solutions are:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Key components:

a = coefficient of $x^2$
b = coefficient of $x$
c = constant term
± = plus or minus (gives us two possible solutions)
$\sqrt{b^2 - 4ac}$ = the discriminant under the square root

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The discriminant ( $b^2 - 4ac$ ) tells us important information about the solutions:

If $b^2 - 4ac > 0$ : Two different real solutions
If $b^2 - 4ac = 0$ : One repeated real solution
If $b^2 - 4ac < 0$ : No real solutions (complex solutions)

Step-by-step method

Identify the coefficients: Write the equation in standard form $ax^2 + bx + c = 0$ and identify values of a, b, and c
Substitute into the formula: Replace a, b, and c in the quadratic formula
Calculate the discriminant: Find $b^2 - 4ac$ under the square root
Simplify: Work out the calculations step by step
Find both solutions: Use the ± to find both possible values of x
Round appropriately: Give answers to the required number of decimal places

Worked example 1

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Worked Example: Solving a quadratic equation with decimal answers

Solve $5x^2 + 7x - 3 = 0$ using the quadratic formula, correct to two decimal places.

Step 1: Identify the coefficients

$a = 5$ , $b = 7$ , $c = -3$

Step 2: Substitute into the formula $x = \frac{-7 \pm \sqrt{7^2 - 4(5)(-3)}}{2(5)}$

Step 3: Calculate the discriminant $x = \frac{-7 \pm \sqrt{49 - (-60)}}{10}$ $x = \frac{-7 \pm \sqrt{49 + 60}}{10}$ $x = \frac{-7 \pm \sqrt{109}}{10}$

Step 4: Simplify $x = \frac{-7 \pm 10.44}{10}$

Step 5: Find both solutions $x = \frac{-7 + 10.44}{10} = \frac{3.44}{10} = 0.34$ $x = \frac{-7 - 10.44}{10} = \frac{-17.44}{10} = -1.74$

Therefore: $x = 0.34$ or $x = -1.74$

Worked example 2

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Worked Example: Solving with exact surd form

Solve $2x^2 - 5x + 1 = 0$ using the quadratic formula.

Step 1: Identify coefficients

$a = 2$ , $b = -5$ , $c = 1$

Step 2: Substitute into formula $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(1)}}{2(2)}$ $x = \frac{5 \pm \sqrt{25 - 8}}{4}$ $x = \frac{5 \pm \sqrt{17}}{4}$

Step 3: Calculate $x = \frac{5 \pm 4.12}{4}$

Step 4: Find solutions $x = \frac{5 + 4.12}{4} = 2.28 \text{ or } x = \frac{5 - 4.12}{4} = 0.22$

Application example - area problems

Real-world applications often require us to use the quadratic formula to solve practical problems involving areas, projectile motion, or optimisation.

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Worked Example: Rectangle area problem

Consider a rectangle with length $(x + 4)$ and width $(x - 1)$ .

If the area is 10 cm², we can set up the equation: $(x + 4)(x - 1) = 10$

Step 1: Expand and rearrange $x^2 + 3x - 4 = 10$ $x^2 + 3x - 14 = 0$

Step 2: Apply quadratic formula Using $a = 1$ , $b = 3$ , $c = -14$ : $x = \frac{-3 \pm \sqrt{9 + 56}}{2} = \frac{-3 \pm \sqrt{65}}{2}$

Step 3: Calculate and interpret Since length cannot be negative, we take the positive solution: $x = \frac{-3 + 8.06}{2} = 2.53 \text{ cm}$

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Common Mistakes and Critical Points:

Always write the equation in standard form first - this helps avoid sign errors
Be careful with negative coefficients - particularly when b is negative
Check your discriminant calculation - errors here affect the entire solution
Remember both solutions - the ± gives two answers unless specifically told otherwise
Round appropriately - follow the question's instructions for decimal places
Verify your answers - substitute back into the original equation when possible

When to use the quadratic formula

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Use the quadratic formula when:

The quadratic expression cannot be factorised easily
You need exact decimal answers
The coefficients are large or awkward numbers
The question specifically asks you to use the formula

The quadratic formula is a universal method that works for any quadratic equation, making it particularly valuable when other methods fail or are impractical.

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Key Points to Remember:

The quadratic formula works for any quadratic equation in the form $ax^2 + bx + c = 0$
Always identify your a, b, and c values correctly before substituting
The ± symbol gives two solutions - don't forget the second one
Check your discriminant ( $b^2 - 4ac$ ) calculation carefully as errors here affect everything
Round to the required number of decimal places as specified in the question
The discriminant tells you about the nature of the solutions before you calculate them

Using the Quadratic Formula (Leaving Cert Mathematics): Revision Notes