This note covers solving quadratic equations by factorising, by the quadratic formula and by completing the square, and solving quadratic inequalities. Quadratics describe projectile paths, areas and many optimisation problems, so being able to solve them quickly and accurately is essential. The formula, completing the square and inequalities are Extended content.
Key ideas
A quadratic equation can be written in the form \(ax^2+bx+c=0\) with \(a\ne0\). It can have two different solutions, one repeated solution or no real solutions. Always rearrange so that one side equals zero before you begin.
Factorising. If the quadratic factorises into \((px+q)(rx+s)=0\), then one bracket must be zero, so \(x=-\dfrac{q}{p}\) or \(x=-\dfrac{s}{r}\). This is the quickest method when it works. An equation like \(x^2=5x\) must not be divided by \(x\): move everything to one side and factorise to \(x(x-5)=0\), so that the solution \(x=0\) is not lost.
The quadratic formula (Extended). This works for every quadratic. The expression \(b^2-4ac\) is called the discriminant. If it is positive there are two real solutions, if it is zero there is one repeated solution, and if it is negative there are no real solutions.
Completing the square (Extended). Write the quadratic as \((x+p)^2+q\). For \(x^2+bx+c\), the number \(p\) is half of \(b\), and \(q=c-p^2\). This form also shows the turning point of the graph at \((-p,q)\). To solve, move the number across, take square roots (remembering \(\pm\)), then subtract \(p\).
Quadratic inequalities (Extended). Solve the matching equation to find the critical values, then sketch the parabola. For a positive \(x^2\) term the curve is a "U" shape.
- Where the curve is below the \(x\)-axis, the quadratic is less than zero. This gives one interval between the roots.
- Where the curve is above the \(x\)-axis, the quadratic is greater than zero. This gives two separate parts, outside the roots.
Write the solution using inequality signs, and use "or" for two separate parts.
Problems in context. Define the unknown, form a quadratic, solve it, and then reject any solution that does not make sense, such as a negative length.
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 \(B=0\). It only works when the right side is zero. |
| \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Quadratic formula, for any quadratic equation (Extended). |
| \(b^2-4ac\) | Discriminant. Positive: two real roots. Zero: one root. Negative: no real roots. |
| \(x^2+bx+c=\left(x+\dfrac b2\right)^2+c-\dfrac{b^2}{4}\) | Completed square form (Extended). |
| Critical values | The roots of the matching equation. They are the boundaries of the solution of an inequality. |
Worked examples
Example 1 (a) Solve \(2x^2-5x-12=0\). (b) A rectangular banner is 5 m longer than it is wide. Its area is 84 m². Find its dimensions.
(a) Multiply the first and last numbers: \(2\times(-12)=-24\). Two numbers that multiply to \(-24\) and add to \(-5\) are \(-8\) and \(3\).
Split the middle term: \(2x^2-8x+3x-12=0\). Factorise in pairs: \(2x(x-4)+3(x-4)=0\).
So \((2x+3)(x-4)=0\), which gives \(x=-\dfrac32\) or \(x=4\).
(b) Let the width be \(w\) metres, so the length is \(w+5\). Then \(w(w+5)=84\).
Rearrange: \(w^2+5w-84=0\). Two numbers with product \(-84\) and sum 5 are 12 and \(-7\), so \((w+12)(w-7)=0\).
A width cannot be negative, so \(w=7\). The length is \(7+5=12\). Check: \(7\times12=84\).
Answer: (a) \(x=-1.5\) or \(x=4\); (b) 7 m by 12 m
Example 2 (Extended) (a) Solve \(3x^2-4x-5=0\) using the formula, giving answers to 3 significant figures. (b) Solve \(x^2+6x-2=0\) by completing the square, giving answers to 3 significant figures.
(a) Here \(a=3\), \(b=-4\), \(c=-5\). The discriminant is \(b^2-4ac=16-4(3)(-5)=16+60=76\).
\[x=\frac{-(-4)\pm\sqrt{76}}{2\times3}=\frac{4\pm\sqrt{76}}{6}\]
Use your calculator for the square root: \(\sqrt{76}=8.7178\ldots\)
So \(x=\dfrac{4+8.7178}{6}=2.12\) or \(x=\dfrac{4-8.7178}{6}=-0.786\).
(b) Half of 6 is 3, so \(x^2+6x=(x+3)^2-9\). The equation becomes \((x+3)^2-9-2=0\), so \((x+3)^2=11\).
Take square roots: \(x+3=\pm\sqrt{11}\), so \(x=-3\pm\sqrt{11}\).
\(\sqrt{11}=3.3166\ldots\), so \(x=0.317\) or \(x=-6.32\).
Answer: (a) \(x=2.12\) or \(x=-0.786\); (b) \(x=0.317\) or \(x=-6.32\)
Example 3 (Extended) Solve \(2x^2-x-10\le0\).
First solve \(2x^2-x-10=0\). Two numbers with product \(-20\) and sum \(-1\) are \(-5\) and \(4\).
\(2x^2-5x+4x-10=0\), so \(x(2x-5)+2(2x-5)=0\), giving \((2x-5)(x+2)=0\).
The critical values are \(x=-2\) and \(x=2.5\).
The \(x^2\) coefficient is positive, so the curve is a "U" shape. It is below or on the \(x\)-axis between the roots.
Test a point inside, \(x=0\): \(-10\le0\) is true, which confirms the middle interval.
Answer: \(-2\le x\le2.5\)
Common errors
- Using the null factor rule when the right-hand side is not zero, such as \((x-2)(x+3)=6\) giving \(x=8\) or \(x=3\). Expand and rearrange to zero first.
- Dividing both sides by \(x\), which loses the solution \(x=0\).
- Wrong signs when substituting into the formula, especially with negative \(b\) (use brackets, \(-(-4)=4\)) and with \(-4ac\) when \(c\) is negative.
- Writing the solution of an "outside" inequality as one combined statement, such as \(3\lt x\lt-3\). Use two parts: \(x\lt-3\) or \(x\gt3\).
- Not rejecting a solution that does not fit the context, such as a negative length.
Exam tip: If a question says "give your answers to 3 significant figures" or "do not use factorising", it is signalling the formula. Write the substituted formula before you use the calculator.
Check yourself
- Solve \(x^2-7x+12=0\).
Answer
\((x-3)(x-4)=0\), so \(x=3\) or \(x=4\). - Solve \(x^2=5x\).
Answer
\(x^2-5x=0\), so \(x(x-5)=0\). Then \(x=0\) or \(x=5\). - (Extended) Solve \(2x^2+3x-1=0\) to 3 s.f.
Answer
The discriminant is \(9+8=17\), so \(x=\dfrac{-3\pm\sqrt{17}}{4}\). This gives \(x=0.281\) or \(x=-1.78\). - (Extended) Solve \(x^2\gt9\).
Answer
The critical values are \(\pm3\). The "U" curve is above the axis outside the roots, so \(x\lt-3\) or \(x\gt3\).