Solving simple quadratic equations

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: $x+2=0\text{or}x+3=0$.

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

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 $x=-3$.