Question

2. i) (2a^2 + 3b)^2
ii) (3x^2 y + z)^2
iii) (2x + 1/3x)^2

Chapter 3 - Algebra - RS Agarwal - ICSE - Class 9th

Answer 1

Hey  there! ☺☻☺

$i) (2a^2 + 3b)^2\\We \ know, (a+b)^2 = a^2 + b^2 + 2ab\\(2a^2 + 3b)^2 = (2a^2)^2 + (3b)^2 + 2 \times 2a^2 \times 3b\\~~~~~~~~~~~~~~~~= 4a^4 + 9b^2 + 12a^2b\\\\ii)(3x^2y + z)^2 = (3x^2y)^2 + (z^2) + 2\times 3x^2y\times z\\~~~~~~~~~~~~~~~~~~~=9x^4y^2 + z^2 + 6x^2yz\\\\iii) \ (2x + \frac{1}{3x}) ^2 = (2x)^2 + (\frac{1}{3x}})^2 + 2 \times 2x + \frac{1}{3x}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~=4x^2 + \frac{1}{9x^2} + \frac{4}{3}$

Answer 2

Hi there!
We'll use the following identity here :

(a + b)² = a² + b² + 2ab

ATQ,

i. (2a² + 3b)²

(2a²)² + (3b)² + 2(2a²)(3b)

4a⁴ + 9b² + 12a² b

ii. (3x² y + z)²

(3x² y)² + (z)² + 2(3x² y)(z)

9x⁴ y2 + z² + 6 x²yz

iii. (2x + 1/3x)²

(2x)² + (1/3x)² + 2(2x)(1/3x)

4x² + 1/9x² + 4/3x²

x² × (4 + 1/9 + 4/3)

x² (49/9)

Hope it helps! :D

Cheers,
Ishaan Singh