Part B: Quadratic Equations and Expressions

Transcript

0:00Introduction

Hello everyone! Okay, one question before we start. Take this equation: x times mod x, minus five mod x, plus six, equals zero. It looks just like a quadratic, doesn't it? So, how many real roots can it have? Two, at most, right? Hmm, keep that answer in mind; we will test it in example one. Today: mixed JEE problems on quadratics, a few quick tools, and the traps examiners love. So, let's begin.

0:32Tool 1: the sum and the product, not the roots

So, tool one. Alpha and beta, these two Greek letters, are the roots of x square, minus a minus two in brackets times x, minus a, minus one, equals zero. Here a belongs to bold R: the symbol like a rounded E means belongs to, and bold R means the real numbers. Find the least value of alpha square plus beta square. Should we find the two roots first? No, no need! The sum, alpha plus beta, is a minus two. The product, alpha beta, is minus, a plus one in brackets. Clear? Now, alpha square plus beta square is alpha plus beta, whole square, minus two alpha beta. Put in the sum and the product: a minus two, whole square, plus two times a plus one. Expand: a square, minus two a, plus six. Complete the square: a minus one, whole square, plus five. The squared part is never negative, so, the double arrow, meaning implies: least value five, at a equals one. So, five. Done? Hmm, toppers always check one more thing. Are these roots real, for every value of a? Check the discriminant, capital D, okay? D equals a minus two, whole square, plus four times a plus one: that is a square plus eight. Always positive, so the roots are real for every a; the tick means checked. Completely safe!

1:58Tool 2: a repeated modulus? Substitute

Alright, tool two. Find the sum of the real roots of: mod of x minus two, whole square, plus mod of x minus two, minus two, equals zero. The vertical bars mean modulus. Now, what repeats here? Yes, mod of x minus two! So call it t, and since a modulus is never negative, t is greater than or equal to zero; that is the sign with a line below. Now it is a simple quadratic: t square plus t minus two equals zero. Factorise: t plus two, times t minus one, equals zero. So t is minus two, or one. Keep both? No, no, careful! Here t is a modulus, so it cannot be minus two. Rejected! Only t equals one survives. Such a favourite trap, isn't it? So mod of x minus two is one. Then x minus two is plus or minus one, this sign on screen. So x is one, or three. Sum of the real roots: one plus three, four. The substitution turned a scary modulus equation into an easy quadratic. Nice and quick!

3:06Topper's trick: complex roots? Go polar

Now a topper's trick, where complex numbers come back. Alpha and beta are the roots of x square minus two x plus four equals zero. Find alpha to the power six, plus beta to the power six. Will you multiply alpha by itself six times? No, please don't! First solve it: the roots are one, plus or minus iota root three. Iota is the imaginary unit, and the root sign means square root. Got it? Now go polar. Tell me: what is the modulus of one plus iota root three, and what is its argument? Yes! Root of one plus three: modulus two. And the argument is pi by three; for the other root, minus pi by three. So the roots are two, e to the power plus or minus iota pi by three. Exactly! Now power six: six times pi by three is two pi. So alpha to the six plus beta to the six is two to the power six, times, in brackets, e to the power iota two pi, plus e to the power minus iota two pi. These two conjugates add up to two cos two pi, so we get two, times two to the power six, times cos two pi; the dot means times. Cos two pi is one, so two times sixty four times one: one hundred and twenty eight. That was fast! Complex roots and high powers? Go polar.

4:31Example 1: number of real roots of x|x| - 5|x| + 6 = 0

Now, example one: our hook question. How many real roots does x mod x, minus five mod x, plus six, equals zero, have? Did you say two? Pause the video and try it first. Hint: two cases for mod x. Done? Okay, let's check. Case one: x greater than or equal to zero, so mod x is x. Then x square minus five x plus six equals zero: x equals two or three. Both are at least zero, so both are fine. Case two: x less than zero, so mod x is minus x. Then the equation becomes minus x square plus five x plus six equals zero. Multiply by minus one: x square minus five x minus six equals zero, so x is six, or minus one. Two roots from each case. So four real roots, right? No, no, careful! Case two needs x less than zero, and six is not. Rejected! Minus one is less than zero, so it stays; the tick means it passes. With me? So the real roots are minus one, two and three. Three real roots! Look, one quadratic gives at most two roots, but this equation is really two quadratics, one for each case. Hook solved!

5:52Example 2: an equation with two square roots

Now a harder one, JEE pattern. Root of three x square minus seven x minus thirty, minus root of two x square minus seven x minus five, equals x minus five. Pause the video and try. Hint: subtract what is under the two roots. Done? Okay, let's check. Short names: capital A for the first root, capital B for the second, okay? So the equation is: A minus B equals x minus five. Now the hint: A square minus B square. The minus seven x cancels, leaving x square minus twenty five: x minus five, times x plus five. And A square minus B square is also A minus B, times A plus B, okay? See, x minus five again! So, divide by A minus B, that is, by x minus five? Any problem with that? Yes! If x equals five, we divide by zero. So test x equals five first. Both expressions under the roots become ten, so A equals B equals root ten, and A minus B is zero, just like five minus five. Tick! So x equals five is a root. Divide blindly, and you lose it: a classic trap! Now x not equal to five; that is the slashed equals sign. Divide: A plus B equals x plus five. Add A minus B equals x minus five: two A equals two x, so A equals x. Neat, isn't it? So root of three x square minus seven x minus thirty equals x, and x is at least zero, since square roots are never negative. Square it: three x square minus seven x minus thirty equals x square, so two x square minus seven x minus thirty equals zero. So x is six, or minus five by two, which is negative: rejected. Last, check x equals six. Capital A is six, capital B is five, and A minus B is one, same as six minus five. Tick! Answer: x equals five, or x equals six. Two roots, and the trap would have eaten one. Got it?

8:04Quick revision: sum and product, modulus, square roots

So, quick revision. One: for symmetric expressions, use the sum and the product; for complex roots, go polar. Two: for a modulus, substitute t, at least zero, or split into cases, and reject roots that break the condition. Three: before dividing by a factor, test the value that makes it zero. Now go and try the practice problems on this page. All the best!