PrepShorts · Study sheet · Class 11 Mathematics · Chapter 4, Complex Numbers and Quadratic Equations
Chapter 4 · Complex Numbers and Quadratic Equations
Square roots of negative numbers, and the surd rule that stops working
This video could not be loaded. Reload the page to try again.
Sign in with Google13 min.
Keep your place in this chapter — sign in, it’s free.Sign in
√(−4) × √(−9) is not √36 — multiplied correctly it is 2i × 3i, which is 6i², minus six. The positive-root rule has nothing to hold once the number turns negative.
The idea
For positive numbers the radical sign picks out one of two candidate roots by choosing the positive one — a tie-break that quietly disappears once the number under the sign is negative, because neither of the two roots is bigger than the other in any sense this chapter defines. So the book has to decide, and it decides that the radical over a negative number shall mean the positive real root times i and nothing else. That decision has a price, and §4.3.6 pays it openly: the school rule that a product of two radicals is the radical of the product now holds only when at most one of the two numbers is negative. The chapter proves the failure rather than warning about it, by showing that keeping the rule would force i² to equal 1.
What you should be able to do
- List both square roots of −1 and verify each by squaring
- Explain why the usual choice of "the positive root" is unavailable for a negative number
- State the convention the chapter adopts for a radical over a negative number
- Write the square roots of a given negative number using that convention, and check them by squaring
- State the range of numbers over which the product-of-radicals rule holds, and the case in which it does not
- Reproduce the contradiction argument that rules out the both-negative case
- Evaluate a product containing a radical over a negative number by converting to the form a + ib first
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| square root | a quantity whose square is the given number | printed in this chapter, §4.3.6 heading, p. 79 |
| negative real number | a real number less than zero — the case §4.3.6 is about | printed in this chapter, §4.3.6 heading, p. 79 |
| positive real number | a real number greater than zero, the case in which the old radical rules were established | printed in this chapter, §4.3.6, p. 79 |
| contradiction | a conclusion that clashes with something already established, used here to discard an assumption | printed in this chapter, §4.3.6, p. 80 |
| convention | a choice the book makes about what a symbol shall mean, not a fact it proves | an added term here; §4.3.6 makes the choice on p. 79 and does not label it on that page |
| principal root | the single value a radical is agreed to name when two candidates exist | an added vocabulary, not printed in this chapter |
| surd rule | the school-level statement that a product of two radicals equals the radical of the product | the explanation's shorthand for the statement worked with on pp. 79–80, which the chapter never names |
Where people slip up
- "The radical over −4 times the radical over −9 is the radical over 36, so 6." This is the error the section exists to kill. Verified: the correct value is (2i)(3i), which is 6i², that is −6. Show both routes side by side and let the sign disagreement do the work.
- "So the product rule for radicals is simply false now." It is not. It holds when both numbers are positive, when exactly one is negative, and when either is zero. Precisely one case fails. Teaching it as "the rule is dead" makes students distrust valid steps.
- "The contradiction shows that i is inconsistent." It shows that the rule, applied where it was never established, is what fails. The chapter's argument begins by assuming the rule and ends by discarding it, not by discarding i.
- "The radical over −1 is ±i." The radical names one value by convention. The equation whose unknown squares to −1 has two solutions. Those are different statements and the page makes both.
- "Because i is a square root, it must be positive or negative." Neither. Nothing in this chapter compares two complex numbers as larger and smaller, so the tie-break that defined the radical over a positive number has nothing to work with here.
- "Multiply first, convert later." Reverse it. Every radical over a negative number should be turned into the form a + ib before any multiplication, which is exactly what Example 4 does in its first line.
Ask your teacher a person
Your teacher reads this and writes back, usually within a day. For an instant answer, use Ask the video in the sidebar.
Your class sees the question and the answer. Only your teacher sees that it was you.
No questions on this topic yet.
Worked answers: Exercise 4.1 · Miscellaneous Exercise
Transcript1,791 words
Two numbers square to minus one. The symbol does, by the relation we started from. And so does its negative, because the two minus signs cancel when you multiply. Two of them. Exactly two, and the search that finds them does not rank them. Now write a square-root sign over minus one and ask what it names. It has to name ONE number. That is what the sign has always done: of the two roots, it picks one and only one.
For a positive number, picking is easy, and everyone learns the rule so early that it stops looking like a rule at all. Take the positive one. Over a negative number, that instruction has nothing to work with, and this video is about what happens next - what gets chosen instead, and what the choice costs. Start with what the sign used to mean, stated carefully enough to be tested.
Four has two square roots, two and minus two. The sign over four names the positive one. That instruction is a tie-break: a test that exactly one of the two candidates passes. Now run the identical test on the two roots of minus four. They are two times the symbol, and minus two times the symbol. Ask each of them whether it is positive, and you get the same answer twice, because neither has anything in the first slot at all - the ordinary part of both of them is empty.
A test that both candidates pass equally is not a tie-break. It does not fail to find a winner because we asked it badly; there is no winner to find. Put the rule to that test across fifteen numbers and the count is flat: it names all seven of the positive ones and none of the seven negative ones. So something else has to do the naming. Here are seven candidate rules, all asked to do the same job.
Take the positive one. Take the one with a positive second slot. Take the one with a negative second slot. Take the one with nothing negative in either slot. Take the one with nothing negative in the first and nothing positive in the second. Take whichever the search happened to meet first. And take anything that squares to the number. That last one is worth a moment. It is true of both roots, so it names neither - it fails to choose by accepting everything, where the schoolroom rule fails by accepting nothing.
Count what each of the seven can name, over the same fifteen numbers. Three of them name a root of every single one. Two of those three are the schoolroom rule extended, one in each of the two obvious directions. Both leave the positive case exactly as it was. And on all seven negative numbers, they name different things. That is the whole situation in one line: where the old rule already decided, every extension agrees; where it had nothing to work with, they part company.
So a decision gets made, and it is worth being blunt about what kind of thing it is. The sign over minus one shall name the symbol. Not its negative. That is a choice about notation, and it is not proved by anything. The other choice was available, is just as consistent, and names the exact negative of what this one names at every negative number - all seven of them.
With the choice made, the general form falls out. The sign over minus two names root two times the symbol. Over minus three, root three times the symbol. Over minus four, twice the symbol. Over minus nine, three times it. And notice what has NOT been decided. The question of which numbers square to minus one still has two answers, and always will. Naming one of them with a symbol does not remove the other. Those are two different statements, and running them together is where most of the confusion in this topic starts.
Now the rule everyone brings with them. The square root of a times the square root of b is the square root of a b. It is a true statement, it is useful, and nearly every student can produce it on demand. It is also a statement that was established for POSITIVE numbers and for nothing else, because until a moment ago the sign was only defined for those. So the honest question is not whether the rule is true. It is: over which numbers is it true, now that the sign means something over more of them?
That is a question you answer by testing, not by remembering. Fifteen numbers, seven negative, one nought, seven positive. Take every ordered pair of them - two hundred and twenty-five pairs - and file each one by the signs of its two numbers. That makes nine cells. For every pair, work out both sides. The product of the two radicals, and the radical of the product. Eight of the nine cells come out completely clean.
Both positive: forty-nine pairs, nothing wrong. One negative and one positive: forty-nine pairs each way, nothing wrong. The cells with a nought in them are clean too, all seven pairs of each, and that is worth saying out loud, because a summary that only warns you about negatives leaves you guessing about nought. One cell fails. Both negative. Forty-nine pairs, and every single one of them is wrong. Not most of them. Not the awkward ones. All forty-nine.
Take one of those forty-nine and do it slowly. The square root of minus four, times the square root of minus nine. Route one, the tempting one: multiply what is under the signs. Minus four times minus nine is thirty-six, and the square root of thirty-six is six. Route two: convert each one first. The square root of minus four is two times the symbol. The square root of minus nine is three times the symbol.
Multiply those and you get six times the symbol squared, which is minus six. Six, and minus six. Both routes are short. Both look like things you have done a hundred times. They disagree, and they disagree by a sign. One of them has to be wrong, and it is worth knowing which, and why. Here is the argument that settles it, and it settles the general case rather than this one example.
Suppose the rule did hold when both numbers are negative. Apply it to the square root of minus one, multiplied by itself. The left-hand side is the symbol times the symbol, which is minus one. That is not in dispute; it is the relation we started from. The right-hand side, if the rule were allowed, would be the square root of minus one times minus one - the square root of one - which is one.
So the supposition forces minus one to be the same number as one. It is not, and there is nothing to negotiate about that. The supposition is therefore false. Notice exactly which supposition: the rule, applied where both numbers are negative. Nothing here casts any doubt on the symbol. The argument uses the symbol perfectly happily on the left-hand side, all the way through. So what went wrong, said plainly?
A statement was established over one range of numbers and then carried outside it. The rule was built on a sign that meant one thing for positives. The sign now also means something for negatives - but by a decision, made a few minutes ago, on grounds that had nothing to do with products. Nobody checked that the decision would keep the old rule working. It was never going to be automatic.
That is a very ordinary kind of mistake, and it is worth naming, because it is not special to square roots. It is also worth saying what the mistake is NOT. The rule is not dead. Eight of the nine cells hold, and they hold for every pair tried in them. Treating it as dead would make a student distrust steps that are perfectly sound. A fair question at this point: was it the choice that broke it? Would the other convention have done better?
No. Run the whole sweep again with the mirror choice - the one that names the negative of everything this one names. Exactly the same cell fails, and the same forty-nine pairs, and each of them off by a sign, just as before. But now try something more drastic: a rule that also changes what the sign over a POSITIVE number means, taking the negative root there too. Something strange happens. The both-negative cell starts working.
And three other cells break instead. So you can move the damage. You cannot make it go away. The two serious candidates each break one cell; the drastic one breaks three, and it costs you the meaning of the square root of four into the bargain. That is the honest shape of it: the failure is not a mistake somebody made. It is the price of extending the sign at all.
Which leaves the practical question. What do you actually do? One instruction, and it is enough: turn every square root of a negative number into the two-slot form BEFORE you multiply anything. Do that and the broken cell never comes up, because you are never applying the rule - you are multiplying two ordinary members of the system. Here is a longer one. Minus root three, plus the square root of minus two, all multiplied by two root three minus the symbol.
First line: convert. The square root of minus two is root two times the symbol. Now multiply the two brackets in the ordinary way. The first slot comes to minus six plus root two. The second comes to root three plus two root six. No rule about products of radicals was used anywhere, which is exactly why nothing could go wrong. One last thing, because it says why anyone bothered.
Long before any of this was tidy, somebody was hunting for two numbers that add to ten and multiply to forty. There are none among the ordinary numbers. Try it and you always fall short. But five plus the square root of minus fifteen, and five minus it, do the job. Add them: the two square-root parts cancel and you get ten. Multiply them: twenty-five plus fifteen, which is forty.
Both conditions, met exactly, by a pair of numbers written with precisely the convention this video has been taking apart. The sign over a negative number is a decision. The decision has a price, and the price is one cell of one rule. It bought a system in which that pair of numbers exists.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- An equation with no real solution, and the symbol invented to solve itClass 11 · Ch 4, Complex Numbers and Quadratic Equations
- Multiplying, then inverting: how division becomes possibleClass 11 · Ch 4, Complex Numbers and Quadratic Equations
Either side of this one
- Why the powers of the new symbol run round a cycle of fourClass 11 · Ch 4, Complex Numbers and Quadratic Equations
- Which school algebra identities survive the enlargement, and whyClass 11 · Ch 4, Complex Numbers and Quadratic Equations