Question

Challenging to all the brainly users who believes that they are studying in reality ......himmat hai toh solve kro

above 9th class students should solve this

options:-
i)x+2y
ii)5x
iii)3x​

Answer 1

Question:-

If 0 < x < 10 and 6 < y < 60, then √(4x² + y² + 4xy) - √(x² + y² - 2xy)

Answer:-

Concept: If a² = b, a = ±√b, not only +√b

Answer: √(4x² + y² + 4xy) - √(x² + y² - 2xy)

= √[ (2x)² + (y)² + 2(2x)(y) ] - √[ (x)² + (y)² - 2(x)(y) ]

= √[ (2x + y)² ] - √[ (x - y)² ]

= ±(2x + y) - [±(x - y)]

First we will find the possible solutions and later verify whether they satisfy the equation or not

Case I :-

In first term positive and in second term also positive is a solution :-

= +(2x + y) - [ +(x - y) ]

= 2x + y - x + y

= x + 2y This is our first solution

Case II :-

In first term positive and in second term negative is a solution :-

= +(2x + y) - [ -(x - y) ]

= 2x + y + x - y

= 3x This is our second solution.

Case III :-

In first term negative and in second term also negative is a solution :-

= -(2x + y) - [ -(x - y) ]

= y - 2x + x - y

= -x This is our third solution

Case IV :-

In first term negative and in second term positive is a solution :-

= -(2x + y) - [ +(x - y) ]

= y - 2x + y - x

= 2y - 3x This is our third solution

So, possible solutions are :-

• x + 2y
• 3x
• -x
• 2y - 3x

Condition Given: 0 < x < 10 and 6 < y < 60

So we need to assume any values of x and y according to the condition (for example, x = 9 and y = 59)

After that we need to see which solution from the above four are satisfying.

For example assuming x = 9 and y = 59, ye will get the value of √(4x² + y² + 4xy) - √(x² + y² - 2xy) as 27.

27 = 3 * 9 = 3x

So, the solution which is following is 3x (when first term is positive and second term is negative)

Ans. 3x => Option (iii)

Answer 2

Answer:

3x

Step-by-step explanation:

Question:-

If 0 < x < 10 and 6 < y < 60, then √(4x² + y² + 4xy) - √(x² + y² - 2xy)

Answer:-

Concept: If a² = b, a = ±√b, not only +√b

Answer: √(4x² + y² + 4xy) - √(x² + y² - 2xy)

= √[ (2x)² + (y)² + 2(2x)(y) ] - √[ (x)² + (y)² - 2(x)(y) ]

= √[ (2x + y)² ] - √[ (x - y)² ]

= ±(2x + y) - [±(x - y)]

First we will find the possible solutions and later verify whether they satisfy the equation or not

Case I :-

In first term positive and in second term also positive is a solution :-

= +(2x + y) - [ +(x - y) ]

= 2x + y - x + y

= x + 2y This is our first solution

Case II :-

In first term positive and in second term negative is a solution :-

= +(2x + y) - [ -(x - y) ]

= 2x + y + x - y

= 3x This is our second solution.

Case III :-

In first term negative and in second term also negative is a solution :-

= -(2x + y) - [ -(x - y) ]

= y - 2x + x - y

= -x This is our third solution

Case IV :-

In first term negative and in second term positive is a solution :-

= -(2x + y) - [ +(x - y) ]

= y - 2x + y - x

= 2y - 3x This is our third solution

So, possible solutions are :-

x + 2y

3x

-x

2y - 3x

Condition Given: 0 < x < 10 and 6 < y < 60

So we need to assume any values of x and y according to the condition (for example, x = 9 and y = 59)

After that we need to see which solution from the above four are satisfying.

For example assuming x = 9 and y = 59, ye will get the value of √(4x² + y² + 4xy) - √(x² + y² - 2xy) as 27.

27 = 3 * 9 = 3x

So, the solution which is following is 3x (when first term is positive and second term is negative)

Ans. 3x => Option (iii)