Alg7.1.3

From Exampleproblems

$x^2\le 4x\,$

Two easy to see answers are $x=0\,$ and $x=4\,$ . Save $x=0\,$ as one solution and assume that $x\ne 0\,$ . You can't divide by $x\,$ because you don't know if it is positive or negative. Assume $x<0\,$ . Now dividing by $x\,$ gives

$x\ge 4\,$

So two numbers greater than and equal to 4 should be tried in place of $x\,$ in the original problem.

Letting $x=4\,$ gives $16\le 16\,$ which is true, but letting $x=5\,$ gives $25\le 20\,$ which is false, so only the number 4 will be kept as a possible answer.

Now assume $x>0\,$ . Dividing by $x\,$ gives

$x \le 4\,$

So two numbers in $(-\infty,0)\,$ and $(0,4)\,$ should be tried in place of $x\,$ in the original problem.

Letting $x=1\,$ gives $1\le 4\,$ but letting $x=-1\,$ gives $1\le -4\,$ which is false, so only the range $(0,4)\,$ will be kept as a possible answer.