How to solve this inequality?

  • Thread starter nightking
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In summary, the conversation discusses how to solve the inequality (a-x+1)(a-x+2) ≤ a, where a is a constant with unknown value. The suggested method is to expand the left hand side and then put it in terms of x. The rules for inequalities are also mentioned, including not dividing by zero and flipping the inequality sign when dividing by a negative number. The final step is to use the difference of two squares formula to find x in terms of a.
  • #1
nightking
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How can I solve this inequality?

(a-x+1)(a-x+2) ≤ a

where a is a constant with unknown value.

Thanks in advance.
 
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  • #2
Hey nightking and welcome to the forums.

You need to expand out the left hand side and then put on side completely in terms of x.

The rules for inequalities are that you can't divide any side by zero (you also have to make sure any variables you have are not zero either if you want to divide), if you divide by a negative number you flip the inequality sign, if you subtract or add a term the sign doesn't change.
 
  • #3
If you want to find x in terms of a, I would start with
((a - x + 3/2) - 1/2)((a - x + 3/2) + 1/2) ≤ a

The left hand side is then the difference of two squares...
 
  • #4
AlephZero said:
If you want to find x in terms of a, I would start with
((a - x + 3/2) - 1/2)((a - x + 3/2) + 1/2) ≤ a

The left hand side is then the difference of two squares...

Brilliant. Thanks!
 
  • #5


To solve this inequality, you will need to follow a few steps. First, distribute the terms in the parentheses to simplify the expression:

a^2 - 2ax + 3a - x^2 + 3x - 2 ≤ a

Next, move all terms to one side of the inequality by subtracting a from both sides:

a^2 - 2ax + 3a - x^2 + 3x - 2 - a ≤ 0

Then, combine like terms and rearrange the equation:

a^2 - x^2 - 2ax + 3x + 2a - a - 2 ≤ 0

a^2 - x^2 - 2ax + 3x + a - 2 ≤ 0

(a-x)^2 + (3x + a - 2) ≤ 0

Now, we can see that the first term, (a-x)^2, will always be greater than or equal to 0. Therefore, for the entire expression to be less than or equal to 0, the second term, (3x + a - 2), must also be less than or equal to 0.

Solving for x in this second term, we get:

3x + a - 2 ≤ 0

3x ≤ -a + 2

x ≤ (-a + 2)/3

Therefore, the solution to this inequality is all values of x that are less than or equal to (-a + 2)/3. This means that any value of x that satisfies this condition will make the original inequality true.

In summary, to solve this inequality, you will need to distribute the terms, move them to one side, combine like terms, and then solve for x. Remember to check your solution by plugging it back into the original inequality to ensure that it is indeed a valid solution.
 

1. What is an inequality?

An inequality is a mathematical statement that compares two quantities using inequality symbols such as <, >, ≤, or ≥. It shows that one quantity is greater or less than the other.

2. How do I solve an inequality?

To solve an inequality, first determine the variable that you are solving for. Then, use algebraic techniques such as adding, subtracting, multiplying, or dividing both sides of the inequality to isolate the variable on one side. Finally, check the solution by plugging it back into the original inequality and making sure it is true.

3. What is the difference between solving an equation and solving an inequality?

In solving an equation, you are finding the value of the variable that makes both sides of the equation equal. In solving an inequality, you are finding the possible values of the variable that make the inequality true.

4. How do I know if my solution is correct?

To check if your solution is correct, plug it back into the original inequality and see if it makes the inequality true. If it does, then your solution is correct. If not, double-check your work for any errors.

5. What are some common mistakes when solving inequalities?

Common mistakes when solving inequalities include forgetting to flip the inequality symbol when multiplying or dividing both sides by a negative number, not distributing correctly when solving multi-step inequalities, and not checking the solution in the original inequality to make sure it is valid.

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