Solving for Delta: √4x-1 - 3 < .5

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In summary, the conversation discusses the use of a graph to find a value for delta in the inequality |√(4x-1)-3| < .5x-2 < delta. The coordinates (2,3) and (3,3.5) are mentioned, but it is unclear how they are related to the problem. The speaker also mentions difficulty with the absolute value of the square root and confusion over using x and X. The problem as given does not seem to be true for values of x around 2, 3, 2.5, or 3.5.
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afcwestwarrior
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| √4x-1 -3 | < .5 X-2<delta

use graph to find delta, u can draw out the graph which is simple, the 4x-1 is squared, it's just i couldn't fit the square root on it all,

now, the coordinates are 2,3 and 3 goes to 3.5 and 2.5

now here's where i get stuck, what do i do with the absolute value of square root 4x-1, because i plug in 3.5 and 2.5 and subtract those with 2 and there not even close to the answer which is 1.44
 
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afcwestwarrior said:
| √4x-1 -3 | < .5 X-2<delta

use graph to find delta, u can draw out the graph which is simple, the 4x-1 is squared, it's just i couldn't fit the square root on it all,
You could at least use parentheses! √(4x-1) or just sqrt(4x-1). Is there a reason for use both x and X? Do they represent different values? And for God's sake don't confuse things by saying "squared" when you mean square root!

now, the coordinates are 2,3 and 3 goes to 3.5 and 2.5
This make no sense at all. The coordinates of WHAT are "2,3" (and do you mean (2, 3) so that an x coordinate is 2 and a y coordinate 3)? I can make no sense at all out of "3 goes to 3.5 and 2.5". 3 is a specific number- it can't "go" to anything else.

now here's where i get stuck, what do i do with the absolute value of square root 4x-1, because i plug in 3.5 and 2.5 and subtract those with 2 and there not even close to the answer which is 1.44
Do you mean putting 3.5 and 2.5 equal to X in .5X- 2?

To begin with, if you actually graph y= √(4x-1)-3 and y= .5x- 2, you will see that √(4x-1)-3< .5x- 2 is not true for x around 2, 3, 2.5, or 3.5. Please state the problem exactly as it is given.
 

1. What is the purpose of solving for delta in this equation?

The purpose of solving for delta in this equation is to find the value of x that satisfies the inequality √4x-1 - 3 < .5. This value of x is known as the critical point and it represents the boundary between the values of x that make the inequality true and those that make it false.

2. How do I solve for delta in this equation?

To solve for delta, you can follow these steps:

  1. Isolate the radical term by adding 3 to both sides of the inequality.
  2. Square both sides of the inequality to eliminate the square root.
  3. Isolate the x term by subtracting 1 from both sides of the inequality.
  4. Divide both sides of the inequality by 4 to isolate x.
  5. The solution for delta is the resulting value of x.

3. Are there any restrictions on the value of x in this equation?

Yes, there are restrictions on the value of x in this equation. Since the square root of a negative number is not a real number, the expression √4x-1 must be greater than or equal to 0. This means that 4x-1 must be greater than or equal to 0. Therefore, the range of possible values for x is restricted to x ≥ 1/4.

4. Can I solve for delta using a graph or table?

Yes, you can use a graph or table to solve for delta in this equation. You can plot the left side of the inequality (√4x-1 - 3) as a function of x and find the x value at which the graph crosses the horizontal line y = .5. Alternatively, you can create a table of x values and plug them into the equation to find the corresponding values for √4x-1 - 3. The value of x at which the resulting value is less than .5 is the solution for delta.

5. What are the practical applications of solving for delta in this equation?

The practical applications of solving for delta in this equation are numerous. This type of inequality can arise in various scientific fields such as physics, engineering, and economics when solving for unknown variables in different types of equations. In these fields, finding the critical point is crucial for making informed decisions and predictions based on the given parameters and constraints.

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