# Tutoring math, you want to optimize your students’ calculator usage. The tutor recalls his own use of the calculator memory back in high school; it was indispensable then.

Let’s imagine you’re facing the following situation:

(^{3}√(25+√14)^2 + (sin(0.27) +2)^3 – ^{3}√(1/7 – √(17))^{2}

True, you could just enter it, start to finish, into a forward entry calculator like the Sharp EL-520W. Assuming you enter everything perfectly, you’ll get the right answer: the scientific calculator will do the order of operations correctly.

With such a long and complex expression, however, the probability of entering something wrong is typically high. One wrong keystroke will result in the wrong answer. What a shame, following so much effort spent trying to enter the expression correctly!

The tutor admits that he’s as likely as anyone to make an entry error. How, then, do I increase the reliability of my entries?

For a long calculation like above, one strategy is to break it into parts, get a reliable answer for each part, then add or subtract them to arrive at the final answer. Let’s revisit our expression from above.

(^{3}√(25)+√(14))^2 + (sin(0.27) +2)^3 – ^{3}√(1/7 – √(17))^{2}

One approach is to get the answer to (^{3}√(25)+√(14))^2. This expression, by itself, is short enough to enter reliably. Even so, I enter it a few times, until I get the same answer over again:

(^{3}√(25)+√(14))^{2}=44.43122487…

I do the same with the second part:

(sin(0.27) +2)^3=11.6466….(calculator in rads)

Similarly with the third part:

^{3}√(1/7 – √(17))^2=2.5115….

Now, I add (or subtract) the three partial answers together to get the final one:

44.4312….+11.6466….-2.5115….=53.56631271

At this point, some of my students will put the brakes on: “My teacher says I’m not allowed to round until the end. How do I get an exact answer if I have to do the computation in stages?”

While you *could* write each partial answer down from the calculator screen, keeping all ten digits each time, then re-enter them to get the final answer, I recognize that’s not practical. Worse yet, it’s downright error-prone: you could (and I likely would) make an error copying the answers down.

What you can do, instead, is use the calculator memory to store the partial answers, then recall them as needed to get the final answer. The key sequence to store an answer from the screen is STO *letterkey*. For instance, to store √(18) in letter E, you’d key in square root 18, then the equals key, then STO, then finally the “log” key (just above the “log” key you’ll see a green ‘E’). To recall that answer later, key in RCL then the “log” key again.

To apply this memory feature to our procedure above: get the answer to (^{3}√(25)+√(14))^{2}, then key in STO CNST to store the answer in the ‘A’ variable. Continue, storing the next partial answer in ‘B’, the last one in C. Now, to get the final answer, key in RCL CNST + RCL y^x (for ‘B’) – RCL x^2 (for ‘C’). You, too, should arrive at 53.56631271.

You can store on top of old values as new calculations demand.

The memory feature can make long calculations much easier and safer. Enjoy playing with it:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

### Like this:

Like Loading...