Quadratic equations

When $a = 1$, find two numbers that multiply to $c$ and add to $b$:

When $a \neq 1$, the reliable route is to split the middle term: find two numbers multiplying to $ac$ and adding to $b$, split, then factorise in pairs. For $6x^2 + 11x - 10$, $ac = -60$ and the pair is $15$ and $-4$:

Two special cases worth spotting instantly: a common factor ($2x^2 - 8x = 2x(x-4)$, roots 0 and 4 — and the root $x = 0$ is the one people lose by dividing through by $x$), and the difference of two squares ($x^2 - 49 = (x-7)(x+7)$).

This is on the 0580 formula list, but write it out before substituting — it is a method mark. Then substitute with brackets around every negative value, because $-b$ with $b = -5$ is $+5$, and that sign is the most common slip in the topic.

The instruction “give your answers correct to 2 decimal places” is a signal: the equation does not factorise, so go straight to the formula. Papers award a mark for a correct substitution even if the arithmetic then goes wrong, so show the substituted formula on its own line before evaluating.

Write $x^2 + bx + c$ as $(x + \tfrac{b}{2})^2 - (\tfrac{b}{2})^2 + c$: halve the coefficient of $x$, square it, subtract it back.

To solve from there: $(x+3)^2 = 11$, so $x + 3 = \pm\sqrt{11}$ and $x = -3 \pm \sqrt{11}$. The $\pm$ is not optional — one square root gives two solutions.

The form also hands you the turning point of the curve $y = x^2 + 6x - 2$, which is $(-3, -11)$: the same question often asks for both.

Extended papers pair one linear equation with one that is not — a quadratic, a circle, or $xy = k$. Elimination will not work; substitution will:

1. Rearrange the linear equation to make one variable the subject.
2. Substitute it into the non-linear one.
3. Rearrange to a quadratic equal to zero, and solve.
4. Substitute each root back into the linear equation to find its partner.
5. Pair the answers: state $x = \ldots, y = \ldots$ and $x = \ldots, y = \ldots$ together.

Step 4 matters: substituting back into the quadratic instead of the linear equation produces extra values that are not solutions of the pair. Step 5 carries its own mark — two loose lists of numbers do not show which $y$ goes with which $x$.

The harder Extended questions do not hand you a quadratic — they describe a situation and expect you to build one. The pattern is almost always a rectangle’s area, a right-angled triangle (Pythagoras), or a journey where time = distance ÷ speed and two times differ by a stated amount.

1. 1.Solving without rearranging to = 0

   $x^2 = 5x + 6$ is not solved by factorising as it stands. Move everything to one side first: $x^2 - 5x - 6 = 0$.

2. 2.Dividing through by x

   $x^2 = 4x$ has roots 0 and 4. Dividing by $x$ destroys the root $x = 0$. Factorise instead: $x(x-4) = 0$.

3. 3.Sign errors in the formula

   Substitute negatives inside brackets: $-(-5) = +5$ and $(-5)^2 = 25$. Write the substitution on its own line before evaluating.

4. 4.One answer from a square root

   $(x+3)^2 = 11$ gives $x + 3 = \pm\sqrt{11}$ — two solutions. The same applies to the $\pm$ in the formula.

5. 5.Substituting back into the non-linear equation

   Put each root into the linear equation. The non-linear one produces extra values that do not satisfy both.

6. 6.Keeping a negative length

   If $x$ is a length, a speed or a number of items, discard the negative root and state that you have — it is a mark on almost every one of these questions.