Question

1)y=|2-√|x||2)y=|x+3|-4​

Answer 1

To find the solutions to the given equations, let's work on them one by one:

Equation 1: y = |2 - √|x| |

Let's break this equation down step by step:

1. Inside the inner absolute value: |x| is always non-negative, so it's never less than 0. Thus, √|x| is real and non-negative.

2. Inside the outer absolute value: 2 - √|x| can also be negative or positive. To cover all possibilities, we'll split this into two cases.

   Case 1: (2 - √|x|) is positive:

   In this case, the absolute value doesn't change the value inside, so we have:

   y = 2 - √|x|

   To isolate x, we can square both sides (since √|x| is non-negative, squaring doesn't change the inequality):

   (y - 2)^2 = |x|

   Now, remember that |a| = a if a is non-negative. So, we can simplify further:

   (y - 2)^2 = x

   Case 2: (2 - √|x|) is negative:

   In this case, the absolute value makes it positive, so we have:

   y = -(2 - √|x|)

   This can be simplified as:

   y = √|x| - 2

   Now, let's square both sides (remember, squaring doesn't change the inequality for non-negative quantities):

   (y + 2)^2 = |x|

   Again, |a| = a if a is non-negative, so we can simplify:

   (y + 2)^2 = x

So, for Equation 1, you have two cases:

1. Case 1: (y - 2)^2 = x
2. Case 2: (y + 2)^2 = x

Equation 2: y = |x + 3| - 4

This equation is a bit simpler:

1. Inside the absolute value: x + 3 can be either positive or negative.

   Case 1: (x + 3) is positive:

   In this case, the absolute value doesn't change the value inside, so we have:

   y = x + 3 - 4

   y = x - 1

   Case 2: (x + 3) is negative:

   In this case, the absolute value makes it positive, so we have:

   y = -(x + 3) - 4

   y = -x - 3 - 4

   y = -x - 7

So, for Equation 2, you have two cases:

1. Case 1: y = x - 1
2. Case 2: y = -x - 7

These are the solutions for the given equations, accounting for all possible cases.