Hi there!
Is a company whose turnover rate is low always excellent? That depends but in the deep-rooted part, there is a trap. Can you guess what it is?
An old cliché goes, “if everybody agrees, there is always someone who doesn’t think.” In a corporate organization, most employees want to feel comfortable working and receive a hefty bonus and salary for just a little work. They don’t want to quit their company, for the entity is a welfare facility for them.
To keep employing such “dangling workers” is a big trap for the management. To talk straight from the gut, the employers had better give them the boot and single out shrewd, lean, and agile employees to get their business more productive.
Say goodbye to those employees who continue working only from sheer force of habit as they are already in an old rut.
Today, let me focus on a trap or loophole as seen in the word problem of the equation.
Solve the equation \(ax + x – a = 0\) for \(x\), providing \(a\) is a real number.
Here is my solution to the problem.
If I solve the equation \(ax + x – a = 0\) for \(x\) as below,
\(ax + x – a = 0\) for \(x\)
\((a + 1)x – a = 0\)
∴ \(x = \frac{a}{a + 1}\) × This is a wrong answer.
In this case, I must solve the equation by diving by cases.
1. Case 1: at \(a = -1\)
\((-1 + 1)x – (-1) = 0\)
\(-x + x + 1 = 0\)
\(1 = 0\) causes contradiction.
2. Case 2: at \(a\neq -1\)
\((a + 1)x = a\)
\(x = \frac{a}{a + 1}\)
From the above Case 1 and Case 2,
There is no answer at \(a = -1\).
\(x = \frac{a}{a + 1}\) at \(a\neq -1\)
Believe it or not, I never fell into a honey trap.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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