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Q2(iv):
Simplify each of the following expressions:
(iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
Solution :
Initial Setup & Given Expression
We are tasked with simplifying the following product of two binomials involving irrational numbers (surds):
$ (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) $
Step 1: Identification of the Algebraic Identity
The given expression perfectly matches the structural form of the Difference of Squares identity. [Per the fundamental theorems of polynomial algebra], the product of the difference and the sum of the same two terms is equal to the square of the first term minus the square of the second term.
The identity is formally stated as:
$ (a - b)(a + b) = a^2 - b^2 $
Step 2: Geometric Verification of the Identity
To rigorously understand why this identity holds true, we can visualize the area of a square of side $a$ from which a smaller square of side $b$ is removed. The remaining area, $a^2 - b^2$, can be geometrically rearranged into a rectangle with dimensions $(a - b)$ and $(a + b)$.
Figure 1: The total shaded area represents $a^2 - b^2$. Factoring out $(a-b)$ from the two shaded rectangles yields $(a-b)(a + b)$.
Step 3: Variable Substitution
By mapping the variables from our specific problem to the algebraic identity, we establish the following equivalences:
- Let $ a = \sqrt{5} $
- Let $ b = \sqrt{2} $
Substituting these values into the right side of the identity ($a^2 - b^2$), we get:
$ (\sqrt{5})^2 - (\sqrt{2})^2 $
Step 4: Application of Exponent Rules on Surds
Next, we evaluate the squares of the square roots. [Per the definition of principal square roots and rational exponents], for any non-negative real number $x$, the operation of squaring its square root returns the original radicand. Mathematically, this is expressed as:
$ (\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x^{\frac{1}{2} \times 2} = x^1 = x \quad \text{for } x \ge 0 $
Applying this axiom to our specific terms:
- $ (\sqrt{5})^2 = 5 $
- $ (\sqrt{2})^2 = 2 $
Step 5: Final Arithmetic Evaluation
Substitute the simplified rational numbers back into the expression from Step 3:
$ 5 - 2 = 3 $
The irrational components have been completely rationalized through the algebraic expansion, resulting in a pure integer.
Final Solution: 3
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.4
- Q1(i): Classify the following numbers as rational or irrational: (i) $2 - \sqrt{5}$
- Q1(ii): Classify the following numbers as rational or irrational: (ii) $(3 + \sqrt{23}) - \sqrt{23}$
- Q1(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- Q1(iv): Classify the following numbers as rational or irrational: (iv) $\frac{1}{\sqrt{2}}$
- Q1(v): Classify the following numbers as rational or irrational: (v) $2\pi$
- Q2(i): Simplify each of the following expressions: (i) $(3 + \sqrt{3})(2 + \sqrt{2})$
- Q2(ii): Simplify each of the following expressions: (ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
- Q2(iii): Simplify each of the following expressions: (iii) $(\sqrt{5} + \sqrt{2})^2$
- Q3: Recall, $\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
- Q4: Represent $\sqrt{9.3}$ on the number line.
- Q5(i): Rationalise the denominators of the following: (i) $\frac{1}{\sqrt{7}}$
- Q5(ii): Rationalise the denominators of the following: (ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
- Q5(iii): Rationalise the denominators of the following: (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
- Q5(iv): Rationalise the denominators of the following: (iv) $\frac{1}{\sqrt{7} - 2}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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