Solving quadratics and completing the squareEdexcel International A Level Maths: Revision notes

A quadratic equation $ax^2+bx+c=0$ can often be solved by factorising into two brackets and using the fact that if a product is zero then at least one factor is zero. Example: $3x^2-7x-6=0$ factorises as $(3x+2)(x-3)=0$, so $3x+2=0$ or $x-3=0$, giving $x=-\frac23$ or $x=3$. Always rearrange to $=0$ first. If the factors multiply to a number other than zero (for example $x(x-3)=10$) you cannot set each factor to that number; expand and rearrange to $x^2-3x-10=0$ before factorising. A root is a solution of the equation, so the equation above has roots $-\frac23$ and $3$.

The solutions of $ax^2+bx+c=0$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ You must know this formula. Use it when the quadratic does not factorise or when the answer is needed as a decimal or a surd. Example (decimals): $3x^2-7x-11=0$ gives $x=\frac{7\pm\sqrt{49+132}}{6}=\frac{7\pm\sqrt{181}}{6}$, so $x=3.41$ or $x=-1.08$ (3 s.f.). Example (exact): $x^2+4x-1=0$ gives $x=\frac{-4\pm\sqrt{20}}{2}=-2\pm\sqrt5$. A calculator's equation solver can check your answers, but you still show the working in the exam.

Completing the square rewrites a quadratic as a perfect square plus a constant: $x^2+bx+c=\left(x+\frac b2\right)^2+\left(c-\frac{b^2}{4}\right).$ Halve the coefficient of $x$, square it inside the bracket, then subtract the same amount to keep the expression equal. Example: $x^2-10x+7=(x-5)^2-25+7=(x-5)^2-18$. To solve $x^2-10x+7=0$: $(x-5)^2=18$, so $x-5=\pm\sqrt{18}=\pm3\sqrt2$ and $x=5\pm3\sqrt2$. This is exact, which is why the method is used when the answer is wanted in surd form. The completed square also shows the minimum value: $(x-5)^2\ge0$ so $f(x)\ge-18$, with equality when $x=5$.

For $ax^2+bx+c$ the general result is $ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right).$ In practice take out the factor $a$ from the $x^2$ and $x$ terms only, complete the square inside, then multiply out. Example: $2x^2+12x+5=2(x^2+6x)+5=2\left[(x+3)^2-9\right]+5=2(x+3)^2-13$. Solving $2x^2+12x+5=0$: $2(x+3)^2=13$, so $(x+3)^2=\frac{13}{2}$ and $x=-3\pm\sqrt{\frac{13}{2}}=-3\pm\frac{\sqrt{26}}{2}$. The quadratic formula gives the same answers. The vertex of the graph is $(-3,-13)$ and the minimum value of the function is $-13$.

• Factorising is quickest when it works: look for integer factors first.
• The formula always works, and is best for decimals.
• Completing the square gives exact surd answers and the vertex or minimum value, and is required when the question says so. A good rule: if the question says 'give your answer in exact form' or 'by completing the square', do not use a decimal from the calculator. If it asks for 3 significant figures, use the formula and round only the final answers. Read the command word: 'solve' needs all values of $x$; 'find the minimum value' needs a $y$-value.

Many problems lead to a quadratic equation. Form the equation, rearrange to $=0$, solve it, then check the answers make sense in the context. Example: a field has width $w$ and length $2w+5$ with area $1375$. Then $w(2w+5)=1375$, so $2w^2+5w-1375=0$ and $(2w+55)(w-25)=0$. The width is $w=25$ m; the root $-27.5$ is rejected because lengths are positive. A path of width $t$ round the outside of the field gives a new area $(25+2t)(55+2t)$. Setting this equal to $2750$ gives $t^2+40t-343.75=0$. Completing the square, $(t+20)^2=743.75$, so $t=-20+\sqrt{743.75}=7.27$ m (3 s.f.). Both sides of the path add $t$, so the length and width each increase by $2t$.