Understand the method
In \(ax^2+bx+c=0\), \(a\ne0\). When \(a=1\), find numbers whose sum is \(b\) and product is \(c\). For a non-unit coefficient, a split middle term or inspection can help.
The zero-product rule applies to an equation with zero on one side. It does not say that a sum of two terms can be solved by setting each term to zero.
Then \(ax+b=0\) or \(cx+d=0\). Some quadratics do not factorise into rational linear factors; completing the square or the quadratic formula can still find their roots.
Three worked examples
Example 1
Solve \(x^2+x-12=0\)
The numbers 4 and -3 have sum 1 and product -12.
- \[(x+4)(x-3)=0\]
Set each factor to zero: \(x=-4\) or \(x=3\). Substitution gives zero in both cases.
The signs in the brackets are the opposite of the corresponding roots.
Example 2
Solve \(6x^2+x-2=0\)
Split the middle term as \(4x-3x\).
- \[2x(3x+2)-(3x+2)=0\]
- \[(3x+2)(2x-1)=0\]
So \(x=-2/3\) or \(x=1/2\). Expanding the factors gives \(6x^2+x-2\).
The leading coefficients multiply to 6; the cross terms must add to x.
Example 3
Solve \(x-\frac4{x-3}=0\)
First record \(x\ne3\). Multiply by \(x-3\), which is nonzero on the allowed domain.
- \[x(x-3)-4=0\]
- \[(x-4)(x+1)=0\]
The candidates are \(4\) and \(-1\). Neither is excluded; substituting each in the original equation confirms both.
Keep restrictions from the original equation, even after the denominator disappears.
Apply and explain
Choose a suitable method
Factorisation is efficient when the factors are easy to find. Otherwise, move to completing the square.
Exam-style check: explain why dividing by x can lose a root
The roots are 0 and 3. Dividing by \(x\) assumes \(x\ne0\), so it loses the valid root 0.
A complete explanation identifies the zero product, gives both roots and states why division excludes one of them.
These study examples explain the reasoning; they are not an official mark scheme.
Common mistakes to catch
- Setting terms of a sum to zero
- Apply the zero-product rule to factors of a product, after rearranging to zero.
- Reporting the bracket constants as roots
- From \(x+4=0\), the root is -4.
- Ignoring a forbidden value
- Check roots against the domain of the original equation.
Now try it yourself
Put the method into practice
In a couple of days, solve \(2x^2-5x-3=0\) again and verify both roots by substitution.