Quadratic Surd

Let $a$ be a positive rational number ($a \in \mathbb{Q}^+$) and $n$ be an integer greater than $1$ ($n \in \mathbb{N}, n \ge 2$). If $a$ is not the $n$-th power of any rational number, then the real root $\sqrt[n]{a}$ (or $a^{1/n}$) is an irrational number. Such an irrational root of a rational number is formally called a surd (করণী) of order $n$.

• When the order $n = 2$, the root is called a quadratic surd (দ্বিঘাত করণী), written simply as $\sqrt{a}$.
• When $n = 3$, it is a cubic surd ($\sqrt[3]{a}$); when $n = 4$, a biquadratic surd ($\sqrt[4]{a}$). In Class 10, the syllabus focuses comprehensively on quadratic surds.
• Critical Criterion: For $\sqrt{a}$ to be a quadratic surd, two conditions MUST hold simultaneously:
  (1) The radicand $a$ must be a positive rational number.
  (2) The value of $\sqrt{a}$ must be strictly irrational.
  Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{10}$ are quadratic surds. But $\sqrt{4} = 2, \sqrt{9} = 3, \sqrt{0.25} = 0.5$ are rational numbers, hence they are NOT surds!

Quadratic surds are categorized based on their rational coefficient structure:

• Pure Quadratic Surd (বিশুদ্ধ দ্বিঘাত করণী): A surd having no rational factor other than $\pm 1$. It contains solely an irrational radical term.
  General form: $\pm \sqrt{a}$, where $a \in \mathbb{Q}^+$ and $a$ is not a perfect square.
  Examples: $\sqrt{2}, -\sqrt{5}, \sqrt{7}, \sqrt{18}$. (Note: $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$; written in single radical form $\sqrt{18}$, it is pure).
• Mixed Quadratic Surd (মিশ্র দ্বিঘাত করণী): A surd having a rational factor other than $\pm 1$, or an expression formed by the sum/difference of a rational number and a pure surd.
  General form: $b\sqrt{a}$ (where $b \in \mathbb{Q}, b \neq 0, \pm 1$) or $a \pm \sqrt{b}$ (binomial surd).
  Examples: $2\sqrt{3}, 5\sqrt{2}, 3 + \sqrt{5}, 7 - 2\sqrt{3}$.

The golden algebraic rule of surd arithmetic states:

• Addition of Like Surds: $a\sqrt{x} + b\sqrt{x} = (a + b)\sqrt{x}$.
  Example: $3\sqrt{2} + 5\sqrt{2} = (3 + 5)\sqrt{2} = 8\sqrt{2}$.
  Example with simplification: $\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$.
• Subtraction of Like Surds: $a\sqrt{x} - b\sqrt{x} = (a - b)\sqrt{x}$.
  Example: $\sqrt{75} - \sqrt{48} = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$.
• Unlike Surds: $\sqrt{2} + \sqrt{3}$ cannot be simplified further. It is a grave error to write $\sqrt{2} + \sqrt{3} = \sqrt{5}$! Notice $\sqrt{2} + \sqrt{3} \approx 1.414 + 1.732 = 3.146$, whereas $\sqrt{5} \approx 2.236$.

For positive rational numbers $a$ and $b$:

• Multiplication Law: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
  For mixed surds: $(p\sqrt{a}) \times (q\sqrt{b}) = (pq)\sqrt{ab}$.
  Examples: $\sqrt{3} \times \sqrt{6} = \sqrt{18} = 3\sqrt{2}$. And $(2\sqrt{5}) \times (3\sqrt{10}) = 6\sqrt{50} = 6 \times 5\sqrt{2} = 30\sqrt{2}$.
• Division Law: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ ($b > 0$).
  Example: $\frac{\sqrt{72}}{\sqrt{8}} = \sqrt{\frac{72}{8}} = \sqrt{9} = 3$ (a rational integer!).
• Distributive Law: $\sqrt{a}(\sqrt{b} + \sqrt{c}) = \sqrt{ab} + \sqrt{ac}$.

If the product of two surds (or an expression containing a surd and another expression) is a rational number, then each is called a rationalising factor (করণী-নিবারক উৎপাদক) of the other. The process of eliminating a surd from an expression by multiplying it with a rationalising factor is called rationalisation (করণী নিরসন).

• For a pure quadratic surd $\sqrt{a}$, the simplest rationalising factor is $\sqrt{a}$ itself, since $\sqrt{a} \times \sqrt{a} = a \in \mathbb{Q}$. (Any non-zero rational multiple $k\sqrt{a}$ is also an RF).
• For a binomial surd $(a + \sqrt{b})$, both $(a - \sqrt{b})$ and $(-a + \sqrt{b})$ are rationalising factors:
  $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b \in \mathbb{Q}$.
  $(a + \sqrt{b})(-a + \sqrt{b}) = (\sqrt{b} + a)(\sqrt{b} - a) = b - a^2 \in \mathbb{Q}$.

While any factor that produces a rational product is a rationalising factor, a conjugate surd is subject to a much stricter mathematical requirement:

Consider the binomial quadratic surd $x = a + \sqrt{b}$ (where $a \in \mathbb{Q}$ and $\sqrt{b}$ is an irrational quadratic surd):

• Candidate 1: $y_1 = a - \sqrt{b}$
  Sum $= (a + \sqrt{b}) + (a - \sqrt{b}) = 2a \in \mathbb{Q}$ (Rational!)
  Product $= (a + \sqrt{b})(a - \sqrt{b}) = a^2 - b \in \mathbb{Q}$ (Rational!)
  Both sum and product are rational $\implies \mathbf{a - \sqrt{b}\text{ is the conjugate surd of } a + \sqrt{b}}$.
• Candidate 2: $y_2 = -a + \sqrt{b}$
  Product $= (a + \sqrt{b})(-a + \sqrt{b}) = b - a^2 \in \mathbb{Q}$ (Product is rational $\implies$ RF).
  Sum $= (a + \sqrt{b}) + (-a + \sqrt{b}) = 2\sqrt{b} \notin \mathbb{Q}$ (Sum is IRRATIONAL!).
  Therefore, $-a + \sqrt{b}$ is a rationalising factor, but it is NOT a conjugate surd!

In standard algebraic form, a fraction is never left with an irrational surd in the denominator. To rationalise the denominator, we multiply both numerator and denominator by the conjugate of the denominator:

1. Monomial Denominator: $$\frac{c}{\sqrt{a}} = \frac{c \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} = \frac{c\sqrt{a}}{a}$$
2. Binomial Denominator: $$\frac{c}{a + \sqrt{b}} = \frac{c(a - \sqrt{b})}{(a + \sqrt{b})(a - \sqrt{b})} = \frac{c(a - \sqrt{b})}{a^2 - b}$$ $$\frac{c}{\sqrt{a} + \sqrt{b}} = \frac{c(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} = \frac{c(\sqrt{a} - \sqrt{b})}{a - b}$$

A high-yield question in WBBSE Madhyamik examinations provides $x = \sqrt{a} + \sqrt{b}$ and $y = \sqrt{a} - \sqrt{b}$ (or $y = 1/x$) and requires determining symmetric polynomial values:

• Base Quantities:
  Sum $x + y = 2\sqrt{a}$
  Difference $x - y = 2\sqrt{b}$
  Product $xy = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b$
• Sum of Squares: $$x^2 + y^2 = (x + y)^2 - 2xy$$
• Sum of Cubes: $$x^3 + y^3 = (x + y)^3 - 3xy(x + y) = (x + y)(x^2 - xy + y^2)$$
• Difference of Cubes: $$x^3 - y^3 = (x - y)^3 + 3xy(x - y) = (x - y)(x^2 + xy + y^2)$$

To find the square root of a binomial surd $x \pm \sqrt{y}$, we express the radicand $x \pm \sqrt{y}$ as the perfect square of another binomial surd $(a \pm \sqrt{b})^2$ or $(\sqrt{a} \pm \sqrt{b})^2$:

1. Expand $(\sqrt{a} \pm \sqrt{b})^2$: $$(\sqrt{a} \pm \sqrt{b})^2 = a + b \pm 2\sqrt{ab}$$
2. Compare this expansion with the given surd $x \pm \sqrt{y}$:
   Equate rational parts: $a + b = x$
   Equate irrational parts: $2\sqrt{ab} = \sqrt{y} \implies 4ab = y \implies ab = \frac{y}{4}$
3. Now find two numbers $a$ and $b$ whose sum is $x$ and whose product is $\frac{y}{4}$.
   Then $\sqrt{x \pm \sqrt{y}} = \pm(\sqrt{a} \pm \sqrt{b})$.