What’s New in Grade 11?#
In Grade 10, you worked with integer exponents. In Grade 11, we extend to:
- Rational exponents (fractional powers like $x^{\frac{2}{3}}$)
- Surds (roots that can’t be simplified to whole numbers, like $\sqrt{3}$)
- Surd equations (equations with square roots)
1. Rational Exponents#
A rational exponent $\frac{m}{n}$ means: “Take the $n$th root, then raise to the power $m$.”
$$ a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m $$Worked Examples#
$8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$
$27^{\frac{4}{3}} = \left(\sqrt[3]{27}\right)^4 = 3^4 = 81$
$16^{-\frac{3}{4}} = \frac{1}{16^{\frac{3}{4}}} = \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8}$
2. Simplifying Surds#
A surd is a root that cannot be simplified to a rational number (e.g., $\sqrt{3}$, $\sqrt{7}$).
The Key Rule#
$$ \sqrt{ab} = \sqrt{a} \times \sqrt{b} $$Use this to simplify by finding the largest perfect square factor:
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
Adding and Subtracting Surds#
You can only combine surds with the same radicand (the number under the root):
$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$ ✓
$3\sqrt{2} + 5\sqrt{3}$ = cannot be simplified ✗
Sometimes you must simplify first#
$\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$
Multiplying Surds#
$\sqrt{3} \times \sqrt{5} = \sqrt{15}$
$2\sqrt{3} \times 4\sqrt{3} = 8 \times 3 = 24$
$(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$ ← Difference of squares!
3. Rationalising the Denominator#
A surd in the denominator is considered “unsimplified.” Multiply top and bottom to remove it.
Type 1: Single surd denominator#
$\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$
Type 2: Binomial denominator (use the conjugate)#
$\frac{4}{3 + \sqrt{2}} = \frac{4}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{4(3 - \sqrt{2})}{9 - 2} = \frac{12 - 4\sqrt{2}}{7}$
The conjugate of $(a + \sqrt{b})$ is $(a - \sqrt{b})$. Multiplying by the conjugate creates a difference of squares, eliminating the surd.
4. Solving Exponential Equations (Grade 11 Level)#
Type 1: Same base#
$3^{2x-1} = 81$
$3^{2x-1} = 3^4$
$2x - 1 = 4 \Rightarrow x = \frac{5}{2}$
Type 2: Quadratic in disguise#
$2^{2x} - 6 \cdot 2^x + 8 = 0$
Let $k = 2^x$:
$k^2 - 6k + 8 = 0$
$(k-2)(k-4) = 0$
$k = 2 \Rightarrow 2^x = 2 \Rightarrow x = 1$
$k = 4 \Rightarrow 2^x = 4 \Rightarrow x = 2$
Type 3: Different bases — use prime factoring#
$6^x = 2^{x+1} \cdot 3^{x-1}$
$(2 \cdot 3)^x = 2^{x+1} \cdot 3^{x-1}$
$2^x \cdot 3^x = 2^{x+1} \cdot 3^{x-1}$
Compare bases: $2^x = 2^{x+1}$ gives a contradiction, so we rearrange:
$\frac{2^x}{2^{x+1}} \cdot \frac{3^x}{3^{x-1}} = 1$
$2^{-1} \cdot 3^{1} = 1 \Rightarrow \frac{3}{2} = 1$? Contradiction — no solution.
5. Solving Surd Equations#
The Method#
- Isolate the surd on one side.
- Square both sides.
- Solve the resulting equation.
- CHECK your answers in the original equation (squaring can create false solutions).
Worked Example 1#
$\sqrt{x + 3} = 5$
Square: $x + 3 = 25$
$x = 22$
Check: $\sqrt{22 + 3} = \sqrt{25} = 5$ ✓
Worked Example 2#
$\sqrt{2x + 1} = x - 1$
Square: $2x + 1 = (x-1)^2 = x^2 - 2x + 1$
$0 = x^2 - 4x$
$0 = x(x - 4)$
$x = 0$ or $x = 4$
Check $x = 0$: $\sqrt{1} = -1$? $1 \neq -1$ ✗ REJECTED
Check $x = 4$: $\sqrt{9} = 3$? $3 = 3$ ✓
Answer: $x = 4$ only.
Worked Example 3 (Two surds)#
$\sqrt{x + 5} - \sqrt{x} = 1$
Isolate one surd: $\sqrt{x + 5} = 1 + \sqrt{x}$
Square: $x + 5 = 1 + 2\sqrt{x} + x$
$4 = 2\sqrt{x}$
$\sqrt{x} = 2$
$x = 4$
Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ ✓
🚨 Common Mistakes#
- Not checking surd equation answers: Squaring can introduce false solutions. ALWAYS substitute back into the original.
- $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$: $\sqrt{9 + 16} = \sqrt{25} = 5$, NOT $3 + 4 = 7$. The root of a SUM is NOT the sum of the roots.
- Rationalising — forgetting to multiply the numerator: When you multiply the denominator by the conjugate, you MUST also multiply the numerator.
- Simplifying $\sqrt{12}$ incorrectly: $\sqrt{12} = 2\sqrt{3}$, NOT $\sqrt{4} \cdot \sqrt{3} = 2 \cdot \sqrt{3}$ written as $2.3$ or $6$. Keep the surd.
- Rejecting valid negative $x$ values: $x$ can be negative in a surd equation as long as the expression UNDER the root is non-negative. $\sqrt{x + 5}$ is valid for $x \geq -5$.
💡 Pro Tip: The “Domain Check”#
Before solving a surd equation, note the domain (what values of $x$ make the expression under the root non-negative). This tells you immediately which answers to reject without substituting.
For $\sqrt{2x + 1} = x - 1$:
- Under the root: $2x + 1 \geq 0 \Rightarrow x \geq -\frac{1}{2}$
- Right side must be non-negative: $x - 1 \geq 0 \Rightarrow x \geq 1$
- So only $x \geq 1$ is valid. This immediately rejects $x = 0$.
🔗 Related Grade 11 topics:
- Quadratic Equations — surd equations often reduce to quadratics after squaring
- The Exponential Graph — exponential equations connect directly to graphing $y = ab^x + q$
📌 Grade 10 foundation: Exponent Laws — the laws you must know before tackling surds
📌 Grade 12 extension: Fundamentals: Exponents — the complete toolkit for matric