This note covers solving quadratic equations by factorising, completing the square and the quadratic formula, and using the discriminant to decide how many real solutions there are. It also shows how to solve quadratic inequalities from a sketch. These skills are used in optimisation, in finding where graphs meet, and in many modelling problems.
Key ideas
A quadratic equation can be written as \(ax^2 + bx + c = 0\) with \(a \neq 0\). Its solutions are also called roots, and they are the \(x\)-intercepts of the graph of \(y = ax^2 + bx + c\). Always rearrange the equation so that one side is 0 before you solve it.
Method 1: factorising. If the quadratic factorises, write it as a product and use the fact that if \(A \times B = 0\), then \(A = 0\) or \(B = 0\). For \(x^2 - 3x - 10 = 0\), write \((x - 5)(x + 2) = 0\), so \(x = 5\) or \(x = -2\).
Method 2: completing the square. Write the quadratic as \(a(x - h)^2 + k = 0\), rearrange to \((x - h)^2 = -\dfrac{k}{a}\), and take square roots, remembering both signs.
Method 3: the quadratic formula. This always works: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Use it when factorising is hard. You can also use the polynomial solver on your GDC to check your answers, or to solve when a calculator method is allowed.
The discriminant. The expression under the square root in the formula is \(\Delta = b^2 - 4ac\). It tells you how many real solutions there are, without solving.
- \(\Delta \gt 0\): two distinct real roots. The graph crosses the \(x\)-axis twice.
- \(\Delta = 0\): one repeated real root. The graph touches the \(x\)-axis at its vertex.
- \(\Delta \lt 0\): no real roots. The graph does not reach the \(x\)-axis.
The discriminant is often used to find unknown values. For example, if a question says an equation has "two equal roots", set \(\Delta = 0\). If it says "real roots", then \(\Delta \geq 0\). If a line and a curve are combined into one quadratic equation, \(\Delta = 0\) means the line is a tangent to the curve, \(\Delta \gt 0\) means it cuts the curve twice, and \(\Delta \lt 0\) means it misses the curve.
Quadratic inequalities. To solve something like \(ax^2 + bx + c \lt 0\):
- Solve the matching equation \(ax^2 + bx + c = 0\) to find the roots.
- Sketch the parabola, showing the roots on the \(x\)-axis.
- Read off where the graph is below the axis (for \(\lt 0\)) or above the axis (for \(\gt 0\)).
For an upward-opening parabola with roots \(p \lt q\), the answer to \(\lt 0\) is \(p \lt x \lt q\), and the answer to \(\gt 0\) is \(x \lt p\) or \(x \gt q\). Include the roots themselves when the inequality has "or equal to".
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(ax^2 + bx + c = 0\) | Standard form of a quadratic equation. Make one side 0 first. |
| \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) | Quadratic formula. Works for every quadratic equation. |
| \(\Delta = b^2 - 4ac\) | The discriminant. |
| \(\Delta \gt 0\) | Two distinct real roots. |
| \(\Delta = 0\) | One repeated real root (tangent to the \(x\)-axis). |
| \(\Delta \lt 0\) | No real roots. |
| Roots to equation | For roots \(p\) and \(q\), the quadratic can be written \(a(x - p)(x - q) = 0\). |
Worked examples
Example 1 Solve \(2x^2 + 3x - 7 = 0\), giving the answers to 3 significant figures.
This does not factorise easily, so use the quadratic formula with \(a = 2\), \(b = 3\), \(c = -7\).
Step 1: discriminant. \(b^2 - 4ac = 9 - 4(2)(-7) = 9 + 56 = 65\).
Step 2: substitute. \(x = \dfrac{-3 \pm \sqrt{65}}{4}\), where \(\sqrt{65} = 8.0623\).
Step 3: the two values. \(x = \dfrac{-3 + 8.0623}{4} = 1.2656\) and \(x = \dfrac{-3 - 8.0623}{4} = -2.7656\).
Answer: \(x = 1.27\) or \(x = -2.77\)
Example 2 Find the values of \(k\) for which the equation \(x^2 + kx + k + 3 = 0\) has two equal real roots.
Step 1: identify \(a = 1\), \(b = k\), \(c = k + 3\).
Step 2: equal roots means \(\Delta = 0\). So \(k^2 - 4(1)(k + 3) = 0\).
Step 3: expand and simplify. \(k^2 - 4k - 12 = 0\).
Step 4: factorise. \((k - 6)(k + 2) = 0\), so \(k = 6\) or \(k = -2\).
Check \(k = 6\): the equation is \(x^2 + 6x + 9 = (x + 3)^2 = 0\), which has a repeated root \(x = -3\). Check \(k = -2\): the equation is \(x^2 - 2x + 1 = (x - 1)^2 = 0\), with repeated root \(x = 1\).
Answer: \(k = 6\) or \(k = -2\)
Example 3 A company model says it makes a loss when \(2x^2 - x - 6 \lt 0\), where \(x\) is the change in the price of a product in euros. Solve the inequality \(2x^2 - x - 6 \lt 0\) to find the price changes that give a loss.
Step 1: solve the matching equation \(2x^2 - x - 6 = 0\). Factorise: \((2x + 3)(x - 2) = 0\), so \(x = -1.5\) or \(x = 2\).
Step 2: sketch. The coefficient of \(x^2\) is positive, so the parabola opens upwards and crosses the \(x\)-axis at \(-1.5\) and \(2\).
Step 3: the graph is below the \(x\)-axis between the roots.
Answer: \(-1.5 \lt x \lt 2\)
Common errors
- Common error: trying to solve before moving every term to one side, for example solving \(x^2 + 3x = 10\) by setting \(x = 10\) or \(x + 3 = 10\). The equation must equal 0 first.
- Common error: dropping a negative sign when substituting into the formula. Put brackets around negative values, such as \((-7)\) for \(c\), and remember that \(-b\) is positive when \(b\) is negative.
- Common error: giving only the positive square root. Both \(+\) and \(-\) must be used.
- Common error: writing a quadratic inequality answer as one inequality when it needs two parts, such as \(x \lt -1\) or \(x \gt 4\) written as \(-1 \gt x \gt 4\), which is impossible.
- Common error: dividing both sides of an equation by \(x\). This loses the solution \(x = 0\). Factorise instead.
Exam tip: when a question mentions the number of roots, use the discriminant straight away. Write down \(a\), \(b\) and \(c\) first, then form \(b^2 - 4ac\).
Check yourself
- How many real solutions does \(3x^2 - 4x + 5 = 0\) have?
Answer
\(\Delta = (-4)^2 - 4(3)(5) = 16 - 60 = -44\). This is negative, so there are no real solutions. - Find the values of \(k\) for which \(x^2 - 6x + k = 0\) has two distinct real roots.
Answer
Need \(\Delta \gt 0\): \(36 - 4k \gt 0\), so \(k \lt 9\). - Solve \(x^2 - 3x - 4 \geq 0\).
Answer
\(x^2 - 3x - 4 = (x - 4)(x + 1)\), with roots \(-1\) and \(4\). The parabola opens upwards, so it is above or on the axis outside the roots: \(x \leq -1\) or \(x \geq 4\). - Find the value of \(k\) for which the line \(y = 2x + k\) is a tangent to the curve \(y = x^2 + 3\).
Answer
Set \(x^2 + 3 = 2x + k\), so \(x^2 - 2x + 3 - k = 0\). A tangent means \(\Delta = 0\): \(4 - 4(3 - k) = 0\), so \(4k - 8 = 0\) and \(k = 2\).