Solving quadratic equations by factorising worksheet (with ... - Free Printable

To solve the given quadratic equations, we will use various methods such as factoring, completing the square, and the quadratic formula. Let's solve each equation step by step.

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Equation 18: \( 4x^2 - 16 = 0 \)

1. Factor out the greatest common factor (GCF):
\[
4(x^2 - 4) = 0
\]

2. Recognize that \( x^2 - 4 \) is a difference of squares:
\[
x^2 - 4 = (x - 2)(x + 2)
\]

3. Substitute back:
\[
4(x - 2)(x + 2) = 0
\]

4. Set each factor equal to zero:
\[
x - 2 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 2 \quad \text{or} \quad x = -2
\]

Solution for Equation 18:
\[
\boxed{x = 2 \text{ or } x = -2}
\]

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Equation 20: \( x^2 + 6x + 9 = 0 \)

1. Notice that \( x^2 + 6x + 9 \) is a perfect square trinomial:
\[
x^2 + 6x + 9 = (x + 3)^2
\]

2. Substitute back:
\[
(x + 3)^2 = 0
\]

3. Solve for \( x \):
\[
x + 3 = 0
\]
\[
x = -3
\]

Solution for Equation 20:
\[
\boxed{x = -3}
\]

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Equation 22: \( x^2 - 3x + 2 = 0 \)

1. Factor the quadratic expression:
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

2. Set each factor equal to zero:
\[
x - 1 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = 1 \quad \text{or} \quad x = 2
\]

Solution for Equation 22:
\[
\boxed{x = 1 \text{ or } x = 2}
\]

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Equation 24: \( 3x^2 + 7x + 2 = 0 \)

1. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3 \), \( b = 7 \), and \( c = 2 \).

2. Calculate the discriminant:
\[
b^2 - 4ac = 7^2 - 4(3)(2) = 49 - 24 = 25
\]

3. Substitute into the quadratic formula:
\[
x = \frac{-7 \pm \sqrt{25}}{2(3)}
\]
\[
x = \frac{-7 \pm 5}{6}
\]

4. Solve for the two roots:
\[
x = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3}
\]
\[
x = \frac{-7 - 5}{6} = \frac{-12}{6} = -2
\]

Solution for Equation 24:
\[
\boxed{x = -\frac{1}{3} \text{ or } x = -2}
\]

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Equation 1: \( x^2 - 10x + 25 = 0 \)

1. Notice that \( x^2 - 10x + 25 \) is a perfect square trinomial:
\[
x^2 - 10x + 25 = (x - 5)^2
\]

2. Substitute back:
\[
(x - 5)^2 = 0
\]

3. Solve for \( x \):
\[
x - 5 = 0
\]
\[
x = 5
\]

Solution for Equation 1:
\[
\boxed{x = 5}
\]

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Equation 3: \( x^2 + 10x + 21 = 0 \)

1. Factor the quadratic expression:
\[
x^2 + 10x + 21 = (x + 3)(x + 7)
\]

2. Set each factor equal to zero:
\[
x + 3 = 0 \quad \text{or} \quad x + 7 = 0
\]
\[
x = -3 \quad \text{or} \quad x = -7
\]

Solution for Equation 3:
\[
\boxed{x = -3 \text{ or } x = -7}
\]

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Equation 5: \( 4x^2 - 2x - 2 = 0 \)

1. Simplify by dividing the entire equation by 2:
\[
2x^2 - x - 1 = 0
\]

2. Factor the quadratic expression:
\[
2x^2 - x - 1 = (2x + 1)(x - 1)
\]

3. Set each factor equal to zero:
\[
2x + 1 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = -\frac{1}{2} \quad \text{or} \quad x = 1
\]

Solution for Equation 5:
\[
\boxed{x = -\frac{1}{2} \text{ or } x = 1}
\]

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Final Answers:

- Equation 18: \( \boxed{x = 2 \text{ or } x = -2} \)
- Equation 20: \( \boxed{x = -3} \)
- Equation 22: \( \boxed{x = 1 \text{ or } x = 2} \)
- Equation 24: \( \boxed{x = -\frac{1}{3} \text{ or } x = -2} \)
- Equation 1: \( \boxed{x = 5} \)
- Equation 3: \( \boxed{x = -3 \text{ or } x = -7} \)
- Equation 5: \( \boxed{x = -\frac{1}{2} \text{ or } x = 1} \)