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Q2(iii):
Simplify each of the following expressions:
(iii) $(\sqrt{5} + \sqrt{2})^2$
Solution :
Given Expression & Algebraic Foundation
We are tasked with simplifying the following mathematical expression involving irrational numbers:
$(\sqrt{5} + \sqrt{2})^2$
To expand this expression systematically, we utilize the standard binomial square identity [Derived from the distributive property of multiplication over addition]:
$(a + b)^2 = a^2 + 2ab + b^2$
Step 1: Variable Substitution
By mapping the given expression to the binomial identity, we establish our variables:
- Let $a = \sqrt{5}$
- Let $b = \sqrt{2}$
Substituting these values into the identity yields:
$(\sqrt{5} + \sqrt{2})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2$
Step 2: Simplifying the Squared Terms
We evaluate the first and third terms of the expansion. [Per the fundamental definition of a principal square root, $(\sqrt{x})^2 = x$ for any non-negative real number $x \ge 0$].
- First term: $(\sqrt{5})^2 = 5$
- Third term: $(\sqrt{2})^2 = 2$
Step 3: Simplifying the Cross-Product Term
Next, we evaluate the middle term, $2(\sqrt{5})(\sqrt{2})$. [Per the Product Property of Radicals, $\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$ for $x, y \ge 0$].
$2(\sqrt{5})(\sqrt{2}) = 2(\sqrt{5 \cdot 2}) = 2\sqrt{10}$
Step 4: Combining Like Terms
Substituting the simplified components back into the expanded equation, we get:
$5 + 2\sqrt{10} + 2$
We group the rational numbers together and leave the irrational term distinct [Per the commutative property of addition]:
$(5 + 2) + 2\sqrt{10} = 7 + 2\sqrt{10}$
Geometric Visualization of the Expansion
The algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ can be visualized as the area of a square with side length $(a+b)$, partitioned into four distinct rectangular regions. Below is the geometric proof applied specifically to $a = \sqrt{5}$ and $b = \sqrt{2}$.
Summing the areas of the four regions confirms our algebraic derivation:
$\text{Total Area} = 5 + \sqrt{10} + \sqrt{10} + 2 = 7 + 2\sqrt{10}$
Final Solution: $7 + 2\sqrt{10}$
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(iv): Simplify each of the following expressions: (iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- 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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