Solving quadratic equationsAQA A-Level Maths: Revision notes

A quadratic equation has the form $ax^2+bx+c=0$. Rearrange so that one side is zero, then factorise and use the fact that if $AB=0$ then $A=0$ or $B=0$. $2x^2-5x-12=0\;\Rightarrow\;(2x+3)(x-4)=0\;\Rightarrow\;x=-\tfrac32\text{ or }x=4.$ A quadratic has up to two roots (solutions). The sum of the roots is $-\frac ba$ and the product is $\frac ca$, a useful check: $4+\left(-\frac32\right)=\frac52=-\frac{-5}{2}$.

When a quadratic does not factorise easily, use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ Write down $a$, $b$, $c$ with their signs and put the numbers in brackets. For $2x^2+8x-3=0$: $x=\frac{-8\pm\sqrt{64+24}}{4}=\frac{-8\pm\sqrt{88}}{4}=\frac{-4\pm\sqrt{22}}{2}$. If a question asks for an exact answer or a surd, leave it in root form. Otherwise give decimals to the accuracy asked for, keeping full calculator values until the end.

Completing the square also solves quadratics, and gives exact answers. For $x^2+bx+c=0$: $\left(x+\frac b2\right)^2=\frac{b^2}{4}-c$, then take the square root, remembering $\pm$. Example: $x^2+4x-5=0\Rightarrow(x+2)^2=9\Rightarrow x=-2\pm3$, so $x=1$ or $x=-5$. If $a\neq1$, take out the factor $a$ first, or divide the whole equation by $a$.

Some equations become quadratics after a substitution. Spot a repeated expression, call it $u$, solve for $u$, then go back to the original variable.

• $x^4-13x^2+36=0$: let $u=x^2$. Then $u^2-13u+36=0$, $u=4$ or $9$, so $x=\pm2$ or $\pm3$.
• $x-7\sqrt x+10=0$: let $u=\sqrt x$. Then $u^2-7u+10=0$, $u=2$ or $5$, so $x=4$ or $25$.
• $2(y+1)^2-5(y+1)-12=0$: let $u=y+1$, so $u=4$ or $-\frac32$ and $y=3$ or $-\frac52$.

In a worded problem, form a quadratic from the information (area, Pythagoras, products) and solve it. Then check each solution against the context. A garden with length $(x+5)$ and width $(x-1)$ has area 72, so $x^2+4x-77=0$, $(x+11)(x-7)=0$. Since $x>1$ is needed for a positive width, reject $x=-11$ and keep $x=7$. If the diagonal is 15 m, then $(x+5)^2+(x-1)^2=225$, which gives $2x^2+8x-199=0$ and $x=8.17$ (the root $-12.2$ is rejected).