Quick questions on Quadratic equations explained: O-Level E-Maths

Once one side is zero and the other is factorised, use the fact that if a product of two factors is zero then at least one factor is zero:

The solutions are the values of the unknown that satisfy the equation, also called the roots. A quadratic usually has two solutions, but it can have one repeated solution (a perfect square such as $x^2 - 6x + 9 = 0$) when both factors are the same.

When the coefficient of $x^2$ is greater than $1$, the reliable factorising method is to "split the middle term". Find two numbers that multiply to $a \times c$ and add to $b$, use them to break the middle term into two, then factorise the four terms in pairs. For $2x^2 - 5x - 3$, the product $ac = -6$ and the numbers $-6$ and $1$ add to $-5$, so $2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$. Checking that the chosen pair multiplies to $ac$ and adds to $b$ before splitting keeps this method dependable on the harder quadratics E-Maths sets.

In a worded problem both algebraic roots are correct, but often only one is physically sensible, and stating which you reject earns a method mark. A length, a time, or a count cannot be negative, so a negative root is discarded. For the rectangle with $x(x + 3) = 40$, the roots are $x = 5$ and $x = -8$; only $x = 5$ is kept because a width cannot be negative. Always solve the quadratic fully first and then apply the context to choose the valid root, rather than discarding a root before solving, which could hide an arithmetic error.

Check that the two numbers both multiply and add correctly, including signs.

Solve $x^2 - 9 = 0$. [2 marks]

Solve $x^2 + 4x = 0$. [2 marks]

Solve $x^2 - 5x - 14 = 0$. [2 marks]