Solving quadratic equations

Topic summary

When solving quadratic equations where the coefficient of \( x^2 \) is greater than 1, the process involves factoring by grouping, and it is slightly more complex than when \( a = 1 \). The general form of such an equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a > 1 \). 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 \( ax^2 + bx + c = 0 \), where \( a \) is greater than 1.

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

Recognise the values of \( a \) (the coefficient of \( x^2 \)), \( b \) (the coefficient of \( x \)), and \( c \) (the constant term) in the quadratic equation.

3. Multiply \( a \) and \( c \):

Multiply the coefficient \( a \) and the constant \( c \) to get the product \( ac \).

4. Find two numbers that multiply to \( ac \) and add to \( b \):

Find two numbers that multiply to \( ac \) and add to \( b \). These two numbers will help split the middle term into two terms.

5. 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 in total.

6. Factor by grouping:

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

7. Solve for \( x \):

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

8. 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 \( 6x^2 + 11x – 35 = 0 \).

1. Write the equation in standard form:

The equation is already in standard form: \( 6x^2 + 11x – 35 = 0 \).

2. Identify \( a = 6 \), \( b = 11 \), and \( c = -35 \):

The values are \( a = 6 \), \( b = 11 \), and \( c = -35 \).

3. Multiply \( a \) and \( c \):

Multiply \( a = 6 \) and \( c = -35 \) to get \( ac = 6 \times -35 = -210 \).

4. Find two numbers that multiply to \( -210 \) and add to \( 11 \):

The numbers \( 21 \) and \( -10 \) work, because \( 21 \times (-10) = -210 \) and \( 21 + (-10) = 11 \).

5. Rewrite the equation:

Rewrite \( 6x^2 + 11x – 35 = 0 \) as \( 6x^2 + 21x – 10x – 35 = 0 \).

6. Factor by grouping:

Group the terms: \( (6x^2 + 21x) + (-10x – 35) = 0 \). Now factor out the common factors: \( 3x(2x + 7) – 5(2x + 7) = 0 \).

7. Factor out the common binomial:

Factor out \( (2x + 7) \): \( (2x + 7)(3x – 5) = 0 \).

8. Solve for \( x \):

Set each binomial equal to zero: \( 2x + 7 = 0 \quad \text{or} \quad 3x – 5 = 0 \).

Solving these gives: \( x = -\frac{7}{2} \quad \text{or} \quad x = \frac{5}{3} \).

Check your solutions:

Substitute \( x = -\frac{7}{2} \) and \( x = \frac{5}{3} \) into the original equation \( 6x^2 + 11x – 35 = 0 \) to confirm they satisfy the equation:

For \( x = -\frac{7}{2} \): \( 6\left(-\frac{7}{2}\right)^2 + 11\left(-\frac{7}{2}\right) – 35 = 6 \times \frac{49}{4} – \frac{77}{2} – 35 = 0 \) (True).

For \( x = \frac{5}{3} \): \( 6\left(\frac{5}{3}\right)^2 + 11\left(\frac{5}{3}\right) – 35 = 6 \times \frac{25}{9} + \frac{55}{3} – 35 = 0 \) (True).

The solutions are:

\( x = -\frac{7}{2} \) and \( x = \frac{5}{3} \).