How to Solve Quadratic Equations using Factorization
A quadratic equation has the general form:
$$ax^2 + bx + c = 0, \quad a \neq 0$$
Factorization means rewriting the quadratic as a product of two brackets, if possible.
Once in factored form, each bracket can be set equal to zero to find the solutions.
Case 1: When the coefficient of \(x^2\) is 1
- Step 1: Identify coefficients \(a\), \(b\), and \(c\).
- Step 2: Find two numbers that multiply to \(c\) and add to \(b\).
- Step 3: Split the middle term and factorize by grouping.
- Step 4: Solve for \(x\) from each factor.
Example 1
Question: Solve \(x^2 - 5x + 6 = 0\).
Step 1: \(a = 1,\; b = -5,\; c = 6\).
Step 2: Numbers are \(-2\) and \(-3\).
Step 3:
$$x^2 - 2x - 3x + 6 = 0$$
$$x(x - 2) - 3(x - 2) = 0$$
$$(x - 2)(x - 3) = 0$$
Step 4: Solutions are \(x = 2\) or \(x = 3\).
Case 2: When the coefficient of \(x^2\) is not 1
Here, we use the “split the middle term” method:
Find two numbers that multiply to \(a \times c\) (product of first and last coefficients)
and add to \(b\). Then split and factorize.
- Step 1: Identify coefficients \(a\), \(b\), \(c\).
- Step 2: Multiply \(a \times c\). Find two numbers that multiply to this product and add to \(b\).
- Step 3: Split the middle term using these numbers and group terms.
- Step 4: Factorize each group and solve for \(x\).
Example 2
Question: Solve \(2x^2 + 5x + 3 = 0\).
Step 1: \(a = 2,\; b = 5,\; c = 3\).
Step 2: Multiply \(a \times c = 2 \times 3 = 6\).
We need two numbers that multiply to 6 and add to 5.
These are 2 and 3.
Step 3: Split the middle term:
$$2x^2 + 2x + 3x + 3 = 0$$
$$(2x^2 + 2x) + (3x + 3) = 0$$
$$2x(x+1) + 3(x+1) = 0$$
$$(2x + 3)(x + 1) = 0$$
Step 4: Solve each factor:
$$2x + 3 = 0 \quad \Rightarrow \quad x = -\tfrac{3}{2}$$
$$x + 1 = 0 \quad \Rightarrow \quad x = -1$$
Final Answer: The solutions are
$$x = -\tfrac{3}{2} \quad \text{or} \quad x = -1$$