Title: Inequalities
Course: Demonstrating STACK (v4.4.7)

Solve the following inequality: \[ [value] \le 0.\]
Write your answer as a logical sequence of inequalities, such as \(5\geq x\) and/or \(x\leq 13\).
Solution :
Feedback: To solve the following inequality: \[ [value] \le 0.\] Factorise to give \(([value])([value]) \le 0.\) The product will be \(\le 0 \mbox{ if }[value] \mbox{ is }\ge 0 \mbox{ and }[value]\le0.\) This solution can be entered as \([value]>=[value] \mbox{ and } [value]<=[value]\).

Solve the following inequality: \[ [value] \le 0.\]
Write your answer as a logical sequence of inequalities, such as "\(x>=13\)" or "\(x>=-9\ \mbox{or}\ x>5\)".

Solution :
Feedback: The factor \(([value])^[value]\) is always positive because of the even power. The other factor \(([value])^[value]\) which has an odd power changes sign across \([value] = [value]\). So the inequality is satisfied for \(([value])\le 0\). This is entered as \([value] <= [value]\)

Solve the following inequality: \[ |[value]| \ge [value].\]
Write your answer as a logical sequence of inequalities, such as \( x>=5\) and/or \(x< 13\).
Feedback: As \[|[value]| \ge [value].\] Then \[[value] \ge [value] \mbox{ or } [value] \le -[value].\] Hence \[[value] \ge [value] +[value] \mbox{ or } [value] \le -[value]+[value].\] \[[value] \ge [value] \mbox{ or } [value] \le [value].\] This can be entered as \([value] <= [value]\) or \([value] >= [value]\).

Solve the following inequality: \[ [value] < [value].\]
Write your answer as a logical sequence of inequalities, such as \( x>=5\) and/or \(x< 13\).
Feedback: As \[|2[value]| < [value].\] Then \[2[value] < [value] \mbox{ and } 2[value] > -[value].\] Hence \[2[value] < [value] + [value] \mbox{ and } 2[value] > -[value]+[value].\] \[2[value] < [value] \mbox{ and } 2[value] > [value].\] \[[value] < [value] \mbox{ and } [value] > 0.\] This can be entered as \([value]>0\) and \([value]<[value]\).

Solve the following inequality: \[ [value]\]
Write your answer as a logical sequence of inequalities, such as "\(x>=13\)" or "\(x>-3\) and \(x<=9\)".
Feedback: The inequality can only be solved where the square root exists. This requires \([value] \mbox{ to be }\ge 0.\) Hence \(-[value] \ge \frac[value][value].\) Therefore the square root exists if \([value] \le [value]\). When this condition is met the inequality is satisfied because the left hand side is positive and the right hand side is negative. This solution is entered as \([value]<=[value]\).

Solve the following inequality: \[ [value] .\]
Write your answer as a logical sequence of inequalities, such as \(x>5\) and/or \(x<=13\).
Feedback: As \[[value].\] Then \[[value] \le [value]\mbox{ and }[value] \ge [value].\] Hence \[[value] \le [value]+[value]\mbox{ and }[value] \ge [value]+[value].\] \[[value] \le [value]\mbox{ and }[value] \ge [value].\] Dividing through by 2 gives \[[value] \le [value]\mbox{ and }[value] \ge [value].\] Taking the 4^th root gives \[[value]\ge -[value]\mbox{ and }[value]\le [value]\mbox{ and }[value] \ge [value]\mbox{ or }[value] \le -[value]\] This is entered as \([value]>=[value] \mbox{ and } [value]<=-1 \mbox{ or }[value]>= 1 \mbox{ and } [value]<=[value]\)