Definition of Rationalising Factor (RF)

The Rationalising Factor (RF) of a given surd is another surd (or expression) which, when multiplied by the given surd, results in a rational number.

Product of two surds is rational, then each of them is called the rationalising factor of the other.

Rationalising Factors for Different Forms of Surds

Key Takeaway: For binomial surds (like $a+\sqrt{b}$ or $\sqrt{a}+\sqrt{b}$), the conjugate surd is the most commonly used and effective rationalising factor.

💡 Rationalising the Denominator

Rationalising the denominator is the process of converting a fraction with an irrational denominator into an equivalent fraction with a rational denominator. This is done by multiplying both the numerator and the denominator by the rationalising factor of the denominator.

Example 1: Rationalising a Denominator with One Surd Term

Solution Steps:

1. Identify the RF: The denominator is $3+\sqrt{2}$. Its conjugate (and RF) is $3-\sqrt{2}$.

2. Multiply: Multiply the fraction by $\frac{3-\sqrt{2}}{3-\sqrt{2}}$ (which is equal to 1):
   $$\frac{1}{3+\sqrt{2}} = \frac{1 \times (3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$$
3. Simplify the Denominator: Use the Difference of Squares formula, $(a+b)(a-b) = a^2 – b^2$:
   $$\frac{3-\sqrt{2}}{(3)^2 – (\sqrt{2})^2} = \frac{3-\sqrt{2}}{9 – 2}$$
4. Final Result:
   $$\frac{3-\sqrt{2}}{7} \quad \text{or} \quad \frac{3}{7} – \frac{\sqrt{2}}{7}$$

Example 2: Rationalising a Denominator with Two Surd Terms

Solution Steps:

1. Identify the RF: The denominator is $\sqrt{8}-\sqrt{7}$. Its conjugate (and RF) is $\sqrt{8}+\sqrt{7}$.

2. Multiply: Multiply the fraction by $\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}$:
   $$\frac{1}{\sqrt{8}-\sqrt{7}} = \frac{1 \times (\sqrt{8}+\sqrt{7})}{(\sqrt{8}-\sqrt{7})(\sqrt{8}+\sqrt{7})}$$
3. Simplify the Denominator: Use the Difference of Squares formula:
   $$\frac{\sqrt{8}+\sqrt{7}}{(\sqrt{8})^2 – (\sqrt{7})^2} = \frac{\sqrt{8}+\sqrt{7}}{8 – 7}$$
4. Final Result:
   $$\frac{\sqrt{8}+\sqrt{7}}{1} = \sqrt{8}+\sqrt{7}$$

   (This can be further simplified since $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. The final simplified answer is $2\sqrt{2}+\sqrt{7}$)

   Example 3: Finding Unknown Coefficients after Rationalisation

   Problem:

   If $\frac{5+\sqrt{3}}{7-4\sqrt{3}} = 47a + \sqrt{3}b$, then find the values of $a$ and $b$.

   Solution Strategy: We must first rationalise the left-hand side (LHS) and then compare the result to the right-hand side (RHS) to equate the rational and irrational parts.

   1. Rationalise the Denominator

   The denominator is $7-4\sqrt{3}$. Its rationalising factor (RF) is the conjugate: $7+4\sqrt{3}$.

   $$\frac{5+\sqrt{3}}{7-4\sqrt{3}} = \frac{(5+\sqrt{3})(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})}$$

   2. Simplify the Numerator (N):

   Use the FOIL method:

   $$N = (5)(7) + (5)(4\sqrt{3}) + (\sqrt{3})(7) + (\sqrt{3})(4\sqrt{3})$$
   $$N = 35 + 20\sqrt{3} + 7\sqrt{3} + 4(\sqrt{3})^2$$
   $$N = 35 + 27\sqrt{3} + 4(3)$$
   $$N = 35 + 27\sqrt{3} + 12$$
   $$N = 47 + 27\sqrt{3}$$

   3. Simplify the Denominator (D):

   Use the Difference of Squares formula, $(A-B)(A+B) = A^2 – B^2$:

   $$D = (7)^2 – (4\sqrt{3})^2$$
   $$D = 49 – (4^2 \times (\sqrt{3})^2)$$
   $$D = 49 – (16 \times 3)$$
   $$D = 49 – 48$$
   $$D = 1$$

   4. Combine and Compare:

   The simplified LHS is:

   $$\frac{47 + 27\sqrt{3}}{1} = 47 + 27\sqrt{3}$$

   We are given:

   $$47 + 27\sqrt{3} = 47a + \sqrt{3}b$$

   By comparing the terms:

   • Rational terms: $47 = 47a \implies a = \frac{47}{47} = 1$

   • Irrational terms: $27\sqrt{3} = \sqrt{3}b \implies b = 27$

   Final Answer: $a=1$ and $b=27$.

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   Example 4: Simplifying a Sum of Rationalised Fractions

   Problem:

   Simplify $\frac{4+\sqrt{5}}{4-\sqrt{5}} + \frac{4-\sqrt{5}}{4+\sqrt{5}}$

   Solution Strategy: Instead of rationalising each fraction separately, combine them first by finding a common denominator.

   1. Combine Fractions

   The common denominator is the product of the two denominators, which are conjugates of each other: $(4-\sqrt{5})(4+\sqrt{5})$.

   $$\text{Expression} = \frac{(4+\sqrt{5})(4+\sqrt{5}) + (4-\sqrt{5})(4-\sqrt{5})}{(4-\sqrt{5})(4+\sqrt{5})}$$
   $$\text{Expression} = \frac{(4+\sqrt{5})^2 + (4-\sqrt{5})^2}{(4)^2 – (\sqrt{5})^2}$$

   2. Simplify the Numerator (N):

   We can use the identity: $(A+B)^2 + (A-B)^2 = 2(A^2 + B^2)$.

   Here, $A=4$ and $B=\sqrt{5}$.

   $$N = 2[(4)^2 + (\sqrt{5})^2]$$
   $$N = 2[16 + 5]$$
   $$N = 2[21]$$
   $$N = 42$$

   (Alternatively, expanding completely: $(16 + 8\sqrt{5} + 5) + (16 – 8\sqrt{5} + 5) = 42)$

   3. Simplify the Denominator (D):

   $$D = (4)^2 – (\sqrt{5})^2$$
   $$D = 16 – 5$$
   $$D = 11$$

   4. Final Result:

   $$\text{Expression} = \frac{42}{11}$$

   Example 5: Proof of a Complex Sum of Rationalised Terms

   Problem:

   Prove that $\frac{1}{3-\sqrt{8}} – \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} – \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-2} = 5$.

   Solution Strategy: Rationalise the denominator of each term individually. The denominators are all conjugates, so the result will be a telescoping series where most terms cancel out.

   1. Rationalise Each Term

   We multiply the numerator and denominator of each fraction by the conjugate of its denominator.

   Term	Conjugate (RF)	Rationalised Denominator (using A2−B2)	Resulting Term
   $\frac{1}{3-\sqrt{8}}$	$3+\sqrt{8}$	$(3)^2 – (\sqrt{8})^2 = 9 – 8 = 1$	$3+\sqrt{8}$
   $\frac{1}{\sqrt{8}-\sqrt{7}}$	$\sqrt{8}+\sqrt{7}$	$(\sqrt{8})^2 – (\sqrt{7})^2 = 8 – 7 = 1$	$\sqrt{8}+\sqrt{7}$
   $\frac{1}{\sqrt{7}-\sqrt{6}}$	$\sqrt{7}+\sqrt{6}$	$(\sqrt{7})^2 – (\sqrt{6})^2 = 7 – 6 = 1$	$\sqrt{7}+\sqrt{6}$
   $\frac{1}{\sqrt{6}-\sqrt{5}}$	$\sqrt{6}+\sqrt{5}$	$(\sqrt{6})^2 – (\sqrt{5})^2 = 6 – 5 = 1$	$\sqrt{6}+\sqrt{5}$
   $\frac{1}{\sqrt{5}-2}$	$\sqrt{5}+2$	$(\sqrt{5})^2 – (2)^2 = 5 – 4 = 1$	$\sqrt{5}+2$

   2. Substitute and Simplify (Telescoping Sum)

   Substitute the rationalised terms back into the expression:

   $$\text{LHS} = (3+\sqrt{8}) – (\sqrt{8}+\sqrt{7}) + (\sqrt{7}+\sqrt{6}) – (\sqrt{6}+\sqrt{5}) + (\sqrt{5}+2)$$

   Distribute the negative signs carefully:

   $$\text{LHS} = 3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2$$

   Cancel the conjugate pairs (the terms ‘telescope’):

   $$\text{LHS} = 3 + (\sqrt{8}-\sqrt{8}) + (-\sqrt{7}+\sqrt{7}) + (\sqrt{6}-\sqrt{6}) + (-\sqrt{5}+\sqrt{5}) + 2$$
   $$\text{LHS} = 3 + 0 + 0 + 0 + 0 + 2$$
   $$\text{LHS} = 5$$

   Since $\text{LHS} = 5$ and $\text{RHS} = 5$, the expression is proved.

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   Example 6: Finding the Value of $x^2 + \frac{1}{x^2}$

   Problem:

   If $x = 2+\sqrt{3}$, find the value of $x^2 + \frac{1}{x^2}$.

   Solution Strategy: Use the algebraic identity $a^2 + b^2 = (a+b)^2 – 2ab$. By substituting $a=x$ and $b=\frac{1}{x}$, we get:

   $$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2\left(x\right)\left(\frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^2 – 2$$

   We first need to calculate the value of $x + \frac{1}{x}$.

   1. Find $\frac{1}{x}$

   $$x = 2+\sqrt{3}$$
   $$\frac{1}{x} = \frac{1}{2+\sqrt{3}}$$

   Rationalise the denominator by multiplying by the conjugate, $2-\sqrt{3}$:

   $$\frac{1}{x} = \frac{1 \times (2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{2-\sqrt{3}}{(2)^2 – (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3}$$
   $$\text{So, } \frac{1}{x} = 2-\sqrt{3}$$

   2. Find $x + \frac{1}{x}$

   $$x + \frac{1}{x} = (2+\sqrt{3}) + (2-\sqrt{3})$$

   The surd terms cancel out:

   $$x + \frac{1}{x} = 2 + 2 + (\sqrt{3}-\sqrt{3}) = 4$$

   3. Calculate $x^2 + \frac{1}{x^2}$

   Use the identity:

   $$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2$$

   Substitute the value found in Step 2:

   $$x^2 + \frac{1}{x^2} = (4)^2 – 2$$
   $$x^2 + \frac{1}{x^2} = 16 – 2$$
   $$x^2 + \frac{1}{x^2} = 14$$

   The final answer is $\mathbf{14}$.

   Example 7: Finding the Value of $x^2 + y^2$ (Image 075c1d, 0760b9)

   Problem: If $x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, find $x^2 + y^2$.

   1. Simplify $x$ (Rationalise the Denominator)

   Multiply $x$ by the conjugate of the denominator, $(\sqrt{3}+\sqrt{2})$:

   $$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$$
   $$x = \frac{3 + 2\sqrt{3\cdot 2} + 2}{3 – 2} = \frac{5 + 2\sqrt{6}}{1}$$
   $$x = 5+2\sqrt{6}$$

   2. Simplify $y$

   Notice that $y$ is the reciprocal of $x$. Multiplying by the conjugate $(\sqrt{3}-\sqrt{2})$:

   $$y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$$
   $$y = \frac{3 – 2\sqrt{3\cdot 2} + 2}{3 – 2} = \frac{5 – 2\sqrt{6}}{1}$$
   $$y = 5-2\sqrt{6}$$

   3. Find $x^2 + y^2$ using the Identity

   Use the identity $x^2 + y^2 = (x+y)^2 – 2xy$. Alternatively, the provided solution uses a variation of the identity: $a^2+b^2 = 2(a_1^2 + b_1^2)$ when $a$ and $b$ are conjugates of the form $a_1+\sqrt{b_1}$ and $a_1-\sqrt{b_1}$.

   • Using the Identity $(A+B)^2 + (A-B)^2 = 2(A^2 + B^2)$:$$x^2 + y^2 = (5+2\sqrt{6})^2 + (5-2\sqrt{6})^2$$

     Here $A=5$ and $B=2\sqrt{6}$.

     $$x^2 + y^2 = 2\left[(5)^2 + (2\sqrt{6})^2\right]$$
     $$x^2 + y^2 = 2\left[25 + (4 \times 6)\right] = 2[25 + 24]$$
     $$x^2 + y^2 = 2[49] = \mathbf{98}$$

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   Example 8: Polynomial Evaluation (Image 075c1d, 0760b9)

   Problem: If $x = \frac{1}{2-\sqrt{3}}$, find the value of $x^3 – 2x^2 – 7x + 5$.

   1. Simplify $x$ (Rationalise the Denominator)

   Multiply $x$ by the conjugate of the denominator, $(2+\sqrt{3})$:

   $$x = \frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2} = \frac{2+\sqrt{3}}{4-3}$$
   $$x = 2+\sqrt{3}$$

   2. Calculate $x^2$ and $x^3$

   • Find $x^2$:
     $$x^2 = (2+\sqrt{3})^2 = (2)^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2$$
     $$x^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$$
   • Find $x^3$:
     $$x^3 = x^2 \cdot x = (7+4\sqrt{3})(2+\sqrt{3})$$
     $$x^3 = 7(2) + 7(\sqrt{3}) + (4\sqrt{3})(2) + (4\sqrt{3})(\sqrt{3})$$
     $$x^3 = 14 + 7\sqrt{3} + 8\sqrt{3} + 4(3)$$
     $$x^3 = 14 + 15\sqrt{3} + 12 = 26 + 15\sqrt{3}$$

   3. Substitute into the Polynomial

   Substitute the values of $x$, $x^2$, and $x^3$ into the expression $x^3 – 2x^2 – 7x + 5$:

   $$x^3 – 2x^2 – 7x + 5$$
   $$= (26 + 15\sqrt{3}) – 2(7 + 4\sqrt{3}) – 7(2 + \sqrt{3}) + 5$$

   4. Simplify the Expression

   Distribute the coefficients:

   $$= 26 + 15\sqrt{3} – (14 + 8\sqrt{3}) – (14 + 7\sqrt{3}) + 5$$
   $$= 26 + 15\sqrt{3} – 14 – 8\sqrt{3} – 14 – 7\sqrt{3} + 5$$

   Group the rational and irrational terms:

   • Rational terms: $(26 – 14 – 14 + 5) = (31 – 28) = 3$

   • Irrational terms: $(15\sqrt{3} – 8\sqrt{3} – 7\sqrt{3}) = (15 – 8 – 7)\sqrt{3} = 0\sqrt{3} = 0$

   $$= 3 + 0 = \mathbf{3}$$

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   This type of polynomial problem is often designed so that $x$ is the root of a simplified quadratic equation.

   • If $x = 2+\sqrt{3}$, then $x-2 = \sqrt{3}$.

   • Squaring both sides: $(x-2)^2 = (\sqrt{3})^2 \implies x^2 – 4x + 4 = 3$.

   • Rearranging: $x^2 – 4x + 1 = 0$.