Solving quadratic equations

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 $a{x}^2+bx+c=0$, 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 $a{x}^2+bx+c=0$, 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: $2x+7=0\text{or}3x-5=0$.

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

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 $x=-\frac{7}{2}$: $6{(-\frac{7}{2})}^2+11(-\frac{7}{2})-35=6\times \frac{49}{4}-\frac{77}{2}-35=0$ (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 $x=\frac{5}{3}$.