- #1

- 2

- 0

(a-x+1)(a-x+2) ≤ a

where a is a constant with unknown value.

Thanks in advance.

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- Thread starter nightking
- Start date

- #1

- 2

- 0

(a-x+1)(a-x+2) ≤ a

where a is a constant with unknown value.

Thanks in advance.

- #2

chiro

Science Advisor

- 4,797

- 133

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

AlephZero

Science Advisor

Homework Helper

- 7,002

- 295

((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

- 2

- 0

((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!

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