Rearranging before solving

Before the zero-product principle can be used, every term must be on one side and \(0\) on the other. Then write the quadratic in the order \(ax^2+bx+c\).

Why: the principle works only for a product equal to \(0\). In \(x(3x+5)=2\), knowing that a product is \(2\) does not tell us either factor, so the \(2\) must first be moved across.

Method

1. Collect all the terms on one side so that the other side is \(0\); keep the \(x^2\)-term positive if you can.
2. Write the terms in descending powers of \(x\) and take out any common factor.
3. Factorise, using a multiplication frame when the coefficient of \(x^2\) is not \(1\).
4. Solve each factor and check the roots in the original equation.

Example

Solve \(3x^2+5x=2\).

\(\begin{aligned} 3x^2+5x-2 &= 0 \\ (3x-1)(x+2) &= 0 \\ x=\dfrac{1}{3} \quad &\text{or}\quad x=-2 \end{aligned}\)

Solve \(12+x=x^2\).

\(\begin{aligned} x^2-x-12 &= 0 \\ (x-4)(x+3) &= 0 \\ x=4 \quad &\text{or}\quad x=-3 \end{aligned}\)

Check: with \(x=\dfrac{1}{3}\), \(3\left(\dfrac{1}{9}\right)+\dfrac{5}{3}=\dfrac{1}{3}+\dfrac{5}{3}=2\). A common error is to move a term without changing its sign; \(12+x=x^2\) becomes \(x^2-x-12=0\), not \(x^2+x+12=0\).