Question

1. i) (4a + 9)^2
ii) (3x + 10y)^2
iii) (_/2 m + _/3 n)^2

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

Answer 1

Given :-

1)

$(4a + 9) {}^{2} \\ \\ = >(4a) {}^{2} + (9) {}^{2} + 2(9 \times 4)a \: \: \: \: \: \\ by \: the \: property \: (a + b) {}^{2} \\ = a {}^{2} + b {}^{2} + 2ab \: \\ \\ = > (4a + 9) {}^{2} \\ = 16a {}^{2} + 81 + 72a$

2)

By using the same property/rule, we get :-

$(3x + 10y) {}^{2} \\ = (3x) {}^{2} + (10x) {}^{2} + 2(3x \times 10y) \\ = 9x {}^{2} + 100y {}^{2} + 60xy$

3)

We get

$( \frac{1}{2m} + \frac{1}{3n} ) {}^{2} \\ = ( \frac{1}{2m} ) {}^{2} + (\frac{1}{3n} ) {}^{2} + 2( \frac{1}{2m} \times \frac{1}{3n} ) \\ = \frac{1}{4m {}^{2} } + \frac{1}{9n {}^{2} } + \frac{1}{3mn}$

Answer 2

Hey mate !!

Here's your answer !!

Refer to the attachment !!

Hope it helps !!

Cheers !!