Question

Chapter:- Expansions.

Solve this question with proper explanation.

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Answer 1

Answer:

(i) 12

________

(ii)-12

________

(iii)5

Step-by-step explanation:

(i) Formula of (a+b)² = a²+b²+2ab

Expanding the R.H.S:

(2x+3)²=4x²+9+12x

= 4x²+12x+9

L.H.S = 4x²+ax+9

Comparing the coefficients :

a = 12

The value of a is equal to 12

________________________________

(ii) Formula of (a-b)² = a²+b²-2ab

Expanding the R.H.S:

(2x-3)²=4x²+9-12x

=4x²-12x+9

L.H.S = 4x²+ax+9

Comparing the coefficients :

a = - 12

The value of a is equal to - 12

________________________________

(iii) Formula of (a+b)² = a²+b²+2ab

Expanding the R.H.S:

(3x+5)²=9x²+25+30x

= 9x²+30x+25

L.H.S = 9x²+(7a-5)x

Comparing the coefficients :

(7a-5)x=30x

7a-5=30

7a=35

a=35/7

a=5

The value of a is equal to 5

________________________________

The above concept is just based on expansion and comparison of the coefficients

Answer 2

$\huge\sf\pink{Answer}$

☞ a = 12

☞ a = -12

☞ a = 5

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ 4x²+ax+9 = (2x+3)²

✭ 4x²+ax+9 = (2x-3)²

✭ 9x²+(7a-5)x+25 = (3x+5)²

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The Value of a?

$\rule{110}1$

$\huge\sf\purple{Steps}$

So in all these sums we first have to expand the LHS and then equate it with the RHS

In Question (1)

➝ $\sf (2x+3)^2 = (2x)^2+2(2x)(3)+3^2$

➝ $\sf 4x^2+12x+9$

Equating with LHS

➝ $\sf \cancel{4x^2}+ax+\cancel{9} = \cancel{4x^2}+12x+\cancel{9}$

➝ $\sf a\cancel{x}=12\cancel{x}$

➝ $\sf\red{a=12}$

In Question (2)

➢ $\sf (2x-3)^2 = (2x)^2 - 2(2x)(3)+3^2$

➢ $\sf 4x^2-12x+9$

Equating with LHS

➢ $\sf\cancel{4x^2}+ax+\cancel{9}=\cancel{4x^2}-12x+\cancel{9}$

➢ $\sf a\cancel{x} = -12\cancel{x}$

➢ $\sf\green{a=-12}$

In Question (3)

➳ $\sf (3x+5)^2 = (3x)^2+2(3x)(5)+5^2$

➳ $\sf 9x^2+30x+25$

Equating with LHS

➳ $\sf \cancel{9x^2} +(7a-5)x+\cancel{25} = \cancel{9x^2}+30x+\cancel{25}$

➳ $\sf (7a-5)\cancel{x} = 30\cancel{x}$

➳ $\sf 7a-5=30$

➳ $\sf 7a=30+5$

➳ $\sf 7a = 35$

➳ $\sf a=\dfrac{35}{7}$

➳ $\sf\orange{a = 5}$

$\rule{170}3$