South African NSC Mathematics

Laws of exponents

For a base $a > 0$ and rational exponents $m$ and $n$, the laws are these.

A rational exponent is a root written as a power: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$. So $8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 4$ and $x^{-\frac{1}{2}} = \dfrac{1}{\sqrt{x}}$. Taking the root first keeps the numbers small.

The laws apply only when the bases are the same, or can be made the same. The sum $2^x + 2^{x+1}$ cannot be combined with the product law, but it can be factorised: $2^x + 2^{x+1} = 2^x(1 + 2) = 3 \cdot 2^x$.

Making the bases the same

Write every base as a product of primes and then collect the powers. To simplify $\dfrac{2^{n+2} \cdot 3^{n}}{6^{n+1}}$:

To simplify $\dfrac{9^{x} \cdot 27^{x-1}}{3^{5x-2}}$, write $9 = 3^2$ and $27 = 3^3$:

When a power has a sum in its exponent and the answer must be a single term, split it with the product law first, for example $5^{x+2} - 5^{x} = 5^x \cdot 25 - 5^x = 5^x(25 - 1) = 24 \cdot 5^x$.

Real numbers

A rational number can be written as $\dfrac{a}{b}$ with $a$ and $b$ integers and $b \neq 0$. Terminating decimals and recurring decimals are rational. An irrational number has a decimal expansion that neither ends nor repeats, such as $\sqrt{2}$, $\pi$ and $\sqrt[3]{5}$. Together they make up the real numbers. A square root of a negative number is not real.

To place an irrational number between two integers, square it against neighbouring perfect squares. Since $64 < 70 < 81$, the number $\sqrt{70}$ lies between 8 and 9.

Surds

A surd is an irrational root, such as $\sqrt{3}$ or $\sqrt[3]{10}$. A root like $\sqrt{16}$ is not a surd because it equals 4, which is rational. The laws for square roots, for $a, b \geq 0$, are

There is no matching law for sums: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. For example $\sqrt{9 + 16} = 5$, while $\sqrt{9} + \sqrt{16} = 7$.

Simplifying

Pull out the largest perfect square factor: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$.

Surds can be added or subtracted only when they are like surds, meaning the same root of the same number, once simplified:

Multiplying a bracket of surds is ordinary expansion, using $(\sqrt{a})^2 = a$:

Rationalising the denominator

A fraction is usually left with a rational denominator. For a single surd, multiply top and bottom by that surd:

For a denominator of the form $a + \sqrt{b}$, multiply by its conjugate $a - \sqrt{b}$. The product is a difference of squares, which removes the root:

Rounding

An answer that is not exact is rounded to the stated number of decimal places, looking at the digit after the last one kept. The value $\sqrt{70} = 8.3666\ldots$ is $8.37$ to two decimal places. Keep full calculator values until the last line so that rounding does not build up.

Equations with surds and rational exponents

To solve an equation with a root in it, isolate the root on one side and square both sides. Squaring can create answers that do not satisfy the original equation, so every solution must be tested in it.

Solve $\sqrt{x + 6} = x$. Squaring gives $x + 6 = x^2$, so $x^2 - x - 6 = 0$ and $(x - 3)(x + 2) = 0$. When $x = 3$, the left side is $\sqrt{9} = 3$, which is correct. When $x = -2$, the left side is $\sqrt{4} = 2$ but the right side is $-2$, so $x = -2$ is rejected. The only solution is $x = 3$.

Solve $x - 1 = \sqrt{2x - 1}$:

A square root is never negative, so $x - 1 \geq 0$ is required. For $x = 2 - \sqrt{2} \approx 0.59$, the value of $x - 1$ is negative, so that root is rejected. For $x = 2 + \sqrt{2}$ the check holds, because $2x - 1 = 3 + 2\sqrt{2} = (1 + \sqrt{2})^2$. The solution is $x = 2 + \sqrt{2}$.

An equation with a rational exponent is solved by raising both sides to the reciprocal power. For $x^{\frac{3}{2}} = 8$, raise both sides to the power $\dfrac{2}{3}$: $x = 8^{\frac{2}{3}} = 4$.