This note covers solving quadratic equations by factorising (both tiers), by the quadratic formula and by completing the square (Higher), and solving linear inequalities (both tiers) and quadratic inequalities (Higher). Quadratics model areas, projectiles and optimisation, so they come up in many problems.
Key ideas
Quadratic equations. A quadratic equation can be written as \(ax^2+bx+c=0\) with \(a\ne0\). It usually has two solutions (also called roots). Always rearrange so that one side is zero before you start.
Solving by factorising. Factorise the expression, then use the fact that if two things multiply to give zero, one of them must be zero. For example, \((x-3)(x+4)=0\) gives \(x=3\) or \(x=-4\). This is the quickest method when the quadratic factorises.
The quadratic formula (Higher tier). For \(ax^2+bx+c=0\), the solutions are given by the formula in the table below. Substitute with care, putting negative numbers in brackets. The part under the root, \(b^2-4ac\), is called the discriminant. If it is positive there are two solutions, if it is zero there is one repeated solution, and if it is negative there are no real solutions.
Completing the square (Higher tier). Write \(x^2+bx+c\) as \((x+p)^2+q\), where \(p=\dfrac{b}{2}\) and \(q=c-p^2\). To solve an equation, rearrange to \((x+p)^2=k\), take square roots of both sides (remember \(\pm\)), and subtract \(p\). It is useful when the question asks for exact answers or gives the form \((x+p)^2+q\).
Linear inequalities. Inequalities use \(\lt\), \(\le\), \(\gt\) and \(\ge\). Solve them like equations, with one important difference: when you multiply or divide both sides by a negative number, reverse the inequality sign. On a number line, use an open circle for \(\lt\) or \(\gt\) and a filled circle for \(\le\) or \(\ge\). A double inequality such as \(-4\lt 2x\le8\) is solved by doing the same operation to all three parts.
Quadratic inequalities (Higher tier). First solve the matching equation to find the critical values. Then sketch the graph of the quadratic, which is a \(\cup\)-shaped curve when the \(x^2\) coefficient is positive. For \(\lt0\) take the part of the \(x\)-axis where the curve is below the axis, which is between the roots. For \(\gt0\) take the part where it is above the axis, which is outside the roots, written as two separate inequalities.
What the roots mean. The solutions of \(ax^2+bx+c=0\) are the \(x\)-coordinates where the graph of \(y=ax^2+bx+c\) crosses the \(x\)-axis. A repeated root means the curve just touches the axis. If the discriminant is negative, the curve does not reach the axis at all, so there are no real solutions.
Choosing a method. At Higher tier, try factorising first, because it is quick and exact. If you cannot see factors within a few seconds, or the question asks for decimal answers to a given accuracy, use the formula. If the question asks for the form \((x+p)^2+q\), or exact answers with surds, use completing the square. Whatever method you use, check each solution in the original equation, using your calculator if needed.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(ax^2+bx+c=0\) | Standard form of a quadratic equation. |
| Null factor rule | If \(AB=0\) then \(A=0\) or \(B=0\). |
| \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Quadratic formula (Higher tier). |
| \(b^2-4ac\) | Discriminant. Positive: 2 roots; zero: 1 root; negative: none. |
| \(x^2+bx+c=\left(x+\dfrac b2\right)^2+c-\dfrac{b^2}{4}\) | Completing the square (Higher tier). |
| Reverse the sign | When multiplying or dividing an inequality by a negative number. |
| Critical values | Roots of the related equation. They are the boundaries of the solution (Higher tier). |
Worked examples
Example 1 A rectangular garden is 3 m longer than it is wide. Its area is 40 m\(^2\). Find the width of the garden.
Let the width be \(x\) metres. Then the length is \((x+3)\) metres.
Area: \(x(x+3)=40\), so \(x^2+3x=40\).
Rearrange to zero: \(x^2+3x-40=0\).
Factorise: two numbers with product \(-40\) and sum \(3\) are \(8\) and \(-5\). So \((x+8)(x-5)=0\).
Solutions: \(x=-8\) or \(x=5\). A width cannot be negative, so reject \(x=-8\).
Check: \(5\times8=40\).
Answer: the width is 5 m (and the length is 8 m).
Example 2 (Higher) (a) Solve \(2x^2+3x-4=0\), giving answers to 3 significant figures. (b) Solve \(x^2+8x+5=0\) by completing the square, giving answers to 3 significant figures.
(a) Here \(a=2\), \(b=3\), \(c=-4\). Substitute into the formula: \(x=\dfrac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)}\).
The discriminant is \(9+32=41\), so \(x=\dfrac{-3\pm\sqrt{41}}{4}\).
With \(\sqrt{41}\approx6.403\): \(x=\dfrac{-3+6.403}{4}=0.851\) or \(x=\dfrac{-3-6.403}{4}=-2.35\).
(b) Half of 8 is 4, so \(x^2+8x+5=(x+4)^2-16+5=(x+4)^2-11\).
Set it to zero: \((x+4)^2=11\), so \(x+4=\pm\sqrt{11}\) and \(x=-4\pm\sqrt{11}\).
With \(\sqrt{11}\approx3.317\): \(x=-0.683\) or \(x=-7.32\).
Answer: (a) \(x=0.851\) or \(x=-2.35\); (b) \(x=-0.683\) or \(x=-7.32\).
Example 3 (a) Solve \(5-2x\ge-3\) and write the positive integer solutions. (b) (Higher) Solve \(x^2-3x-10\lt0\).
(a) Subtract 5 from both sides: \(-2x\ge-8\).
Divide both sides by \(-2\) and reverse the sign: \(x\le4\).
The positive integers that satisfy this are 1, 2, 3 and 4.
(b) Solve the equation first: \(x^2-3x-10=(x-5)(x+2)=0\), so the critical values are \(x=5\) and \(x=-2\).
The graph of \(y=x^2-3x-10\) is a \(\cup\) shape crossing the \(x\)-axis at \(-2\) and \(5\). It is below the axis between these values.
Answer: (a) \(x\le4\); the integers are 1, 2, 3, 4. (b) \(-2\lt x\lt5\).
Common errors
- Writing \(x(x+3)=40\) and then claiming \(x=40\) or \(x+3=40\). The null factor rule only works when the other side is zero, so move everything to one side first.
- Forgetting to reverse the inequality when dividing by a negative number.
- Missing the minus sign in the formula: the first term is \(-b\), so for \(b=-5\) it is \(+5\).
- Giving only one solution to a quadratic, or not rejecting a solution that does not make sense in the context, such as a negative length.
- For \(\gt0\) quadratic inequalities, writing a single combined statement such as \(-3\lt x\lt5\) instead of two separate regions.
Exam tip: for a quadratic inequality, always draw a quick sketch of the curve and shade the part of the axis you need. It stops you from choosing the wrong region.
Check yourself
- Solve \(x^2-5x+6=0\).
Answer
\((x-2)(x-3)=0\), so \(x=2\) or \(x=3\). - Solve \(3-4x\lt11\).
Answer
\(-4x\lt8\). Divide by \(-4\) and reverse the sign: \(x\gt-2\). - (Higher) Solve \(x^2+4x-3=0\), giving answers to 3 significant figures.
Answer
\(x=\dfrac{-4\pm\sqrt{16+12}}{2}=\dfrac{-4\pm\sqrt{28}}{2}\). With \(\sqrt{28}\approx5.292\): \(x=0.646\) or \(x=-4.65\). - (Higher) Solve \(x^2-2x-15\gt0\).
Answer
\((x-5)(x+3)=0\) gives critical values \(-3\) and \(5\). The \(\cup\)-shaped curve is above the axis outside these values, so \(x\lt-3\) or \(x\gt5\).