IBM 610 Manual Of Operation

Also see for 610: Technical overview and introduction

BASIC ARITHMETIC OPERATIONS 17The product is in the A register, and the A registeris selected. We multiply by c in register 03, recallingthat the product is in A, with A selected:X 03We can combine these two steps on one line20 X 01 X 03and have the answer (4ac) in A, with A selected.We wish to store 4ac for later use in a storage reg-ister, say 1 1:11 COPY AWe next wish to develop b". We recall that b standsin storage register 02. Hence:02 X 02gives b^ in the A register, with A selected.Because we want b^ — 4ac, we subtract 4ac (in 11)from register A:- 11The difference stands in the A register, with A se-lected.We now wish to take the square root of the differ-ence. We notice that if we take the square root ofthe difference by simply depressing the V key,we would destroy the difference (b^ — 4ac) and re-place it by Vb^ — 4ac. Care must be exercised not todestroy any intermediate results that -may be neededlater. In this case, we do not need to retain the dif-ference, b^ — 4ac; so we depress the "v/ key and getVb* — 4ac in the A register, with the A register se-lected.Because we have completed usage of the constant(-f-b) for this problem, it would be convenient tochange it to — b. This is accomplished by:02 CNVAt this point, we notice that both values of x aredeveloped in almost the same manner, the exceptionbeing the sign of the radical. Let us now store — b inone additional register for use in developing the nu-merators of the two fractions. We do this by selectinga register (09) and putting — b in it:09 COPY 0209 now stands selected, and \/b^ — 4ac is in A:The key depressions:+Agives -b +\/b^ - 4ac in 09.To develop -b -Vb^ - 4ac in 02, we instruct themachine:02 - AWe are now ready to divide the two developed nu-merators by 2 a in order to obtain the two values of x.However, we note that we have only the value of ana in location 01. In order to obtain 2a, we might in-sert a constant 2 into an arbitrary storage register byan ENT instruction and multiply. However, it iseasier to merely say:01 -f 01This would give 2 a in 01 ; the value of the a would bedestroyed in the process.Now we are ready to divide numerator by denomi-nator to obtain our values of x. For the first value:02 -f- 01The quotient stands in A, with A selected. Let us storethe first value of x momentarily:15 COPY ANow we are ready to obtain the second value of x:09 H- 01The quotient stands in A, with A selected (Figure 4).+ A oq Develop -b+U^'i'ae. /n n^is1k> 09 ^02. — A 02. Develop -b-U^-¥cu. intt01 + 01 01 Deifelop 2* fh heqisfe*- 01oa. ^ 01 A Uevelof ^b-Yb'-Zae, />, A ypquT^Y2.IK/r COPY A IS- Sfohe, -L- yi>'--4aju /h yeaisfty IS" »v»eiit*ia.of -;- Ol A Develop -h +U^-¥eu^ /* A t-fiif«>2*1 JFigure 4. Answer Development