Solving simple quadratic equations

Topic summary

When solving quadratic equations where the coefficient of \( x^2 \) is 1, the process is straightforward. The general form of such an equation is \( x^2 + bx + c = 0 \), where \( b \) and \( c \) are constants. The goal is to factorise the quadratic expression and solve for \( x \). Here is a step-by-step guide to solving these types of equations.

1. Write the quadratic equation in standard form:

Ensure the quadratic equation is written as \( x^2 + bx + c = 0 \), where the coefficient of \( x^2 \) is 1.

2. Identify the values of \( b \) and \( c \):

Recognise the values of \( b \) (the coefficient of \( x \)) and \( c \) (the constant term) in the quadratic equation.

3. Factorise the quadratic expression:

Find two numbers that multiply to \( c \) and add to \( b \). These two numbers will be used to split the middle term.

4. Rewrite the equation by splitting the middle term:

Rewrite the equation by expressing \( bx \) as two terms, using the numbers found in the previous step. This will give you four terms.

5. Factor by grouping:

Group the terms into two pairs and factor out the common factor from each pair. You should now be left with two common binomial factors.

6. Solve for \( x \):

Set each binomial factor equal to zero and solve for \( x \).

7. Check your solutions:

Substitute the values of \( x \) back into the original equation to verify that they satisfy the equation.

Example

Solve the quadratic equation \( x^2 + 5x + 6 = 0 \).

1. Write the equation in standard form:

The equation is already in standard form: \( x^2 + 5x + 6 = 0 \).

2. Identify \( b = 5 \) and \( c = 6 \):

The values are \( b = 5 \) and \( c = 6 \).

3. Find two numbers that multiply to \( 6 \) and add to \( 5 \):

The numbers \( 2 \) and \( 3 \) work, because \( 2 \times 3 = 6 \) and \( 2 + 3 = 5 \).

4. Rewrite the equation:

Rewrite \( x^2 + 5x + 6 = 0 \) as \( x^2 + 2x + 3x + 6 = 0 \).

5. Factor by grouping:

Group the terms: \( (x^2 + 2x) + (3x + 6) = 0 \). Now factor out the common factors: \( x(x + 2) + 3(x + 2) = 0 \).

6. Factor out the common binomial:

Factor out \( (x + 2) \): \( (x + 2)(x + 3) = 0 \).

7. Solve for \( x \):

Set each binomial equal to zero: \( x + 2 = 0 \quad \text{or} \quad x + 3 = 0 \).

Solving these gives: \( x = -2 \quad \text{or} \quad x = -3 \).

Check your solutions:

Substitute \( x = -2 \) and \( x = -3 \) into the original equation \( x^2 + 5x + 6 = 0 \) to confirm they satisfy the equation:

For \( x = -2 \): \( (-2)^2 + 5(-2) + 6 = 4 – 10 + 6 = 0 \) (True).

For \( x = -3 \): \( (-3)^2 + 5(-3) + 6 = 9 – 15 + 6 = 0 \) (True).

The solutions are:

\( x = -2 \) and \( x = -3 \).