Oxford PAT (Physics Aptitude Test)

## Quadratic equations A quadratic $ax^2+bx+c=0$ with $a\neq 0$ can be solved by factorising, by completing the square or with the formula. $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ The expression under the root, $b^2-4ac$, is the **discriminant**. When it is positive there are two distinct real roots, when it is zero there is one repeated root (the curve touches the $x$-axis at its turning point) and when it is negative there are no real roots. The roots add up to $-b/a$ and multiply to $c/a$, which is often quicker than solving when a question only asks about their sum or product. For $2x^2-5x-3=0$ the discriminant is $25+24=49$, so $x=(5\pm7)/4$, giving $x=3$ or $x=-\tfrac12$. The sum is $2.5=5/2$ and the product is $-1.5=-3/2$, as the rules predict. ### Completing the square Write $x^2+bx$ as $\left(x+\frac b2\right)^2-\frac{b^2}{4}$. For $x^2+6x+1$ this gives $(x+3)^2-9+1=(x+3)^2-8$, so the smallest value of the expression is $-8$, reached at $x=-3$. When the coefficient of $x^2$ is not 1, take it out first: $2x^2-8x+5=2(x-2)^2-3$. ### Equations that are quadratics in disguise An equation such as $x^4-5x^2+4=0$ becomes a quadratic after the substitution $u=x^2$. Then $u^2-5u+4=0$ gives $u=1$ or $u=4$, so $x=\pm1$ or $x=\pm2$. Every root for $u$ must be converted back, and any negative value of $u$ gives no real $x$. ## Indices For $a\neq0$ the laws of indices are $$ a^m\times a^n=a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad (a^m)^n=a^{mn},\qquad a^0=1,\qquad…