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.

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 fractional power is a root: $a^{1/n}$ is the $n$th root of $a$, and $a^{m/n}=\left(a^{1/n}\right)^m$. So $8^{2/3}=2^2=4$ and $27^{-1/3}=1/3$. The laws only combine powers of the same base, so $2^3\times 4^2$ is first rewritten as $2^3\times 2^4=2^7=128$.

Surds

A surd is a root that cannot be written as a fraction, such as $\sqrt2$ or $\sqrt{12}=2\sqrt3$. The rules $\sqrt{ab}=\sqrt a\sqrt b$ and $\sqrt{a/b}=\sqrt a/\sqrt b$ hold for positive $a$ and $b$, but $\sqrt{a+b}$ is not $\sqrt a+\sqrt b$.

To rationalise a denominator of the form $1/\sqrt a$, multiply top and bottom by $\sqrt a$ to get $\sqrt a/a$. For a denominator such as $p+\sqrt q$, multiply by the conjugate $p-\sqrt q$, because $(p+\sqrt q)(p-\sqrt q)=p^2-q$ has no surd.

As a check, both sides are about $0.708$.

Simultaneous equations

For one linear and one quadratic equation, rearrange the linear one to make a variable the subject and substitute it into the quadratic. With $y=x+1$ and $x^2+y^2=13$:

So $x=2$ with $y=3$, or $x=-3$ with $y=-2$. Always substitute back into the linear equation, because using the quadratic can produce pairs that do not belong together.

Three equations in three unknowns are handled the same way: eliminate one unknown at a time until a single equation in one variable is left. With $y=x^2$, $z=x+y$ and $z+y=6$, substituting gives $x+2x^2=6$, so $2x^2+x-6=0$ and $x=\tfrac32$ or $x=-2$. The corresponding values are $(x,y,z)=(1.5,\,2.25,\,3.75)$ and $(-2,\,4,\,2)$, and both satisfy $z+y=6$.

Polynomials: the factor and remainder theorems

When a polynomial $f(x)$ is divided by $(x-a)$, the remainder is $f(a)$ (the remainder theorem). In particular $(x-a)$ is a factor of $f(x)$ exactly when $f(a)=0$ (the factor theorem).

For $f(x)=x^3+2x-5$ divided by $x-2$ the remainder is $f(2)=8+4-5=7$. For $f(x)=x^3-2x^2-5x+6$, trying small integers that divide the constant term gives $f(1)=1-2-5+6=0$, so $(x-1)$ is a factor. Dividing,

so the roots are $1$, $3$ and $-2$. Any rational root $p/q$ of a polynomial with integer coefficients has $p$ dividing the constant term and $q$ dividing the leading coefficient, which gives a short list of candidates to test.

Binomial expansion

For a positive integer $n$,

The coefficients are a row of Pascal's triangle: $(1+x)^5=1+5x+10x^2+10x^3+5x^4+x^5$. To find one term, use the general term directly. The coefficient of $x^2$ in $(2+3x)^4$ is $\binom42\times2^2\times3^2=6\times4\times9=216$. Take care to raise the number in front of $x$ to the same power as $x$.

Inequalities

Inequalities are solved like equations except that multiplying or dividing both sides by a negative number reverses the sign. From $-2x>6$ it follows that $x<-3$.

For a quadratic inequality, find the roots and sketch the parabola, then read off where it is above or below the axis. To solve $15+2x-x^2>0$, rewrite it as $x^2-2x-15<0$, which factorises as $(x-5)(x+3)<0$. An upward parabola is below the axis between its roots, so $-3<x<5$. Never divide by an expression containing $x$ unless its sign is known, since the sign may be negative for part of the range.

Modulus

The modulus $|x|$ of a real number is its distance from zero, so it is never negative: $|x|=x$ for $x\ge0$ and $|x|=-x$ for $x<0$. An equation such as $|2x+1|=5$ splits into $2x+1=5$ and $2x+1=-5$, giving $x=2$ or $x=-3$. An inequality $|x-2|<3$ means the distance from $2$ is less than $3$, so $-1<x<5$, while $|x-2|>3$ means $x<-1$ or $x>5$.