Solving the Inequality: How to Find the Solution for (a-x+1)(a-x+2) ≤ a?

[nightking]

How can I solve this inequality?

(a-x+1)(a-x+2) ≤ a

where a is a constant with unknown value.

Thanks in advance.

 

[chiro]

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.

 

[AlephZero]

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...

 

[nightking]

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!

 

[CellCoree]

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.