Question

1) 16-b^2= 2) (11x+2)^2= 3) (5a-6b)^2= 4) (3-5a)^3=

Answer 1

1) (16 - b)^2 = 256 - 32b + b^2
2) (11x+2)^2 = 11x^2 + 44x + 4
3) (5a-6b)^2 = 25a^2 - 60ab + 36b^2
4) (3-5a)^3 = 27 - 135a + 225a^2 - 125a^3

Answer 2

Let's solve each of these equations step by step:

1. 16 - b^2 = 0

To solve this quadratic equation, you can rearrange it and factor it as follows:

b^2 = 16

Now, take the square root of both sides:

b = ±√16

b = ±4

So, the solutions for this equation are b = 4 and b = -4.

2. (11x + 2)^2 = 0

To solve this quadratic equation, you can take the square root of both sides:

11x + 2 = 0

Now, isolate x:

11x = -2

Divide by 11:

x = -2/11

So, the solution for this equation is x = -2/11.

3. (5a - 6b)^2 = 0

To solve this equation, take the square root of both sides:

5a - 6b = 0

Now, isolate a:

5a = 6b

Divide by 5:

a = (6b)/5

So, the solution for this equation is a = (6b)/5.

4. (3 - 5a)^3 = 0

To solve this equation, take the cube root of both sides:

3 - 5a = 0

Now, isolate a:

5a = 3

Divide by 5:

a = 3/5

So, the solution for this equation is a = 3/5.