Strand 4 — Algebra

Quadratic Equations

3rd Year (Junior Cert)

✓By the end of this lesson students will be able to identify a quadratic equation and write it in standard form.
✓By the end of this lesson students will be able to solve quadratic equations by factorisation.
✓By the end of this lesson students will be able to solve quadratic equations using the quadratic formula.
✓By the end of this lesson students will be able to verify solutions to quadratic equations by substitution.

Key concepts

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term where the variable is squared, but no term where the variable is raised to a higher power. The standard form of a quadratic equation is ax² + bx + c = 0, where 'x' is the variable, and 'a', 'b', and 'c' are constants, with 'a' not equal to zero. The solutions to a quadratic equation are also called its roots.

ax² + bx + c = 0 (where a ≠ 0)

Solving by Factorisation

To solve a quadratic equation by factorisation, you first ensure the equation is in standard form (ax² + bx + c = 0). Then, factorise the quadratic expression into two linear factors. The 'Zero Product Rule' states that if the product of two or more factors is zero, then at least one of the factors must be zero. By setting each factor equal to zero, you can find the values of 'x' that satisfy the equation.

If (x - p)(x - q) = 0, then x - p = 0 or x - q = 0.

Solving by the Quadratic Formula

The quadratic formula is a general method for finding the solutions (roots) of any quadratic equation in the form ax² + bx + c = 0. It is particularly useful when the quadratic expression cannot be easily factorised. The formula provides the values of 'x' directly.

x = [-b ± √(b² - 4ac)] / 2a

Key facts to remember

1A quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.
2The solutions to a quadratic equation are also known as its roots.
3A quadratic equation can have up to two distinct real solutions.
4To solve by factorisation, use the Zero Product Rule: if (x - p)(x - q) = 0, then x = p or x = q.
5The quadratic formula is x = [-b ± √(b² - 4ac)] / 2a.
6Always ensure the quadratic equation is set equal to zero before attempting to solve it.
7The term b² - 4ac inside the square root is called the discriminant; it tells us about the nature of the roots.

Worked examples

Example 1

Solve the quadratic equation x² + 7x + 10 = 0 by factorisation.

IThe equation is already in standard form: x² + 7x + 10 = 0.

IIWe need to find two numbers that multiply to 10 and add to 7. These numbers are 2 and 5.

IIIFactorise the quadratic expression: (x + 2)(x + 5) = 0.

IVApply the Zero Product Rule: Set each factor equal to zero.

Vx + 2 = 0 OR x + 5 = 0

VISolve each linear equation:

VIIx = -2 OR x = -5

Answer

x = -2, x = -5

Always check your solutions by substituting them back into the original equation.

Example 2

Solve the quadratic equation 2x² - 5x - 3 = 0 by factorisation.

IThe equation is in standard form: 2x² - 5x - 3 = 0.

IIWe need to factorise the trinomial 2x² - 5x - 3. We look for factors of 2x² (2x and x) and factors of -3 (e.g., 1 and -3, -1 and 3).

IIIBy trial and error (or grouping method), we find the factors:

IV(2x + 1)(x - 3) = 0.

VApply the Zero Product Rule: Set each factor equal to zero.

VI2x + 1 = 0 OR x - 3 = 0

VIISolve each linear equation:

VIII2x = -1 => x = -1/2

9x = 3

Answer

x = -1/2, x = 3

Factorising trinomials where a ≠ 1 often requires more careful thought or the 'splitting the middle term' method.

Example 3

Solve the quadratic equation x² - 4x + 1 = 0 using the quadratic formula. Give your answers correct to two decimal places.

IIdentify the values of a, b, and c from the standard form ax² + bx + c = 0.

IIFor x² - 4x + 1 = 0, we have a = 1, b = -4, c = 1.

IIIWrite down the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a.

IVSubstitute the values of a, b, and c into the formula:

Vx = [-(-4) ± √((-4)² - 4(1)(1))] / 2(1)

VISimplify the expression:

VIIx = [4 ± √(16 - 4)] / 2

VIIIx = [4 ± √12] / 2

9Calculate the two possible values for x:

10x₁ = (4 + √12) / 2 AND x₂ = (4 - √12) / 2

11Using a calculator, √12 ≈ 3.4641...

12x₁ = (4 + 3.4641) / 2 = 7.4641 / 2 = 3.73205...

13x₂ = (4 - 3.4641) / 2 = 0.5359 / 2 = 0.26795...

14Round to two decimal places:

15x₁ ≈ 3.73

16x₂ ≈ 0.27

Answer

x ≈ 3.73, x ≈ 0.27

The quadratic formula always works, even when factorisation is difficult or impossible with integer factors. Pay close attention to signs when substituting values into the formula.

Common mistakes

✗Not setting the equation to zero before attempting to factorise or apply the formula.
✗Incorrectly applying the Zero Product Rule (e.g., if (x+2)(x+5) = 10, then x+2=10 or x+5=10, which is incorrect).
✗Making sign errors when substituting values into the quadratic formula, especially with negative 'b' values.
✗Calculation errors, particularly with the square root or division in the quadratic formula.
✗Forgetting to find both solutions (the ± part) when using the quadratic formula.

Exam tips

★Always show all steps in your working for full marks, even if you can do some steps in your head.
★If a question asks for answers to a certain number of decimal places, use the quadratic formula and round at the very end.
★Practise both factorisation and the quadratic formula regularly to become proficient in both methods.
★After finding solutions, quickly substitute them back into the original equation to verify your answers, especially for factorisation problems.

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