Quadratic Expressions in Two Variables

Transcript

0:00Introduction

Hello everyone! First, one small puzzle. Take x square, minus five x y, plus six y square. It splits easily: x minus two y, times x minus three y. Now add four x, minus fourteen y, plus lambda. For which lambda does it still split into two linear factors? Trial and error? That could take forever! Today one neat condition settles it. And trust me, this idea is useful for JEE preparation. So, let's begin.

0:35Second-degree expression in x and y

So, here is the most general expression of degree two in x and y: a x square, plus two h x y, plus b y square, plus two g x, plus two f y, plus c. The six letters are constants. And notice the little two with h, g and f. Keep that in mind, okay? Now the big question. When can this be written as a product of two linear factors, brackets like something x, plus something y, plus a constant? Where would you even start? Hmm, tough one, right? Here's one simple move. Treat y as a constant, and read it as a quadratic in x: a x square, plus two times the bracket h y plus g, times x, plus the bracket b y square plus two f y plus c. Now y lives inside the coefficients. Clear? Now solve for x with the quadratic formula. x equals minus the bracket h y plus g, plus or minus the square root of h y plus g whole square, minus a times the bracket b y square plus two f y plus c, and all of this divided by a. Here we take a not equal to zero, okay? For linear factors, x must come out linear in y, okay? The first part, minus the bracket h y plus g, over a, is already linear. But the square root? When does it become linear in y? Yes, exactly! When the part under the root is a perfect square in y. Then the square root opens up, and x is linear in y. This double-headed arrow means if and only if, okay? The part under the root is a quadratic in y, and it is a perfect square exactly when its own discriminant, in y, is zero. Simplify that, and we get one famous condition. So, let's see it.

2:30The condition: Δ=0

So here is the result. The expression is a product of two linear factors, real or complex, if and only if delta equals zero. This triangle-shaped symbol is capital delta, a Greek letter, okay? And delta is: a b c, plus two f g h, minus a f square, minus b g square, minus c h square. Write this down; it is the heart of today's lesson. Got it? Too many letters? Here is a topper's trick. It is simply a three by three determinant; the two straight bars mean determinant. Row one: a, h, g. Row two: h, b, f. Row three: g, f, c, okay? So a, b, c run down the diagonal, and h, g, f are mirrored across it. Really easy to remember.

3:24Topper's trick: the range of one variable

Alright, now a second idea that toppers love. Suppose x and y are real, linked by one quadratic relation. We want the range of y, that is, all the values y can take. So, a quadratic in which variable? We want y, so we make it a quadratic in y, obviously. Right? No, no, careful! Treat it as a quadratic in the other variable, x. Then y sits inside the coefficients, just like before. Got it? And x must be real, right? So the discriminant, capital D, is greater than or equal to zero. The double arrow means implies. That inequality in y alone gives the range of y. That's it, that's the trick!

4:09Example 1: greatest value of y when x and y are real

So, let's try the trick. x and y are real, and x square minus two x y plus two y square minus two y minus three equals zero. Find the greatest value of y. Pause the video, and try it first. Done? Okay, let's check. We want y, so read it as a quadratic in x. x square, minus two y times x, plus the bracket two y square minus two y minus three, equals zero. Clear? Now x is real, so the discriminant is greater than or equal to zero. The middle coefficient is minus two y, so we use D over four: half of it, squared, minus the constant term. That is, y square minus the bracket two y square minus two y minus three, greater than or equal to zero. Open the bracket: minus y square, plus two y, plus three, greater than or equal to zero. I want y square to be positive, so I multiply by minus one. Quick question: what happens to the greater than or equal to sign? Yes, it flips! Multiplying by a negative number flips the inequality. So y square minus two y minus three is less than or equal to zero, that is, y minus three, times y plus one, less than or equal to zero. Forgetting this flip is a classic JEE trap, so watch out! So y lies from minus one to three, both included. The greatest value of y is three. See? One discriminant, and done. Super fast.

5:49Example 2: find lambda for linear factors, then factorise

Now the harder one, and guess what, it is our puzzle from the start. Find lambda so that x square minus five x y plus six y square plus four x minus fourteen y plus lambda splits into linear factors, and factorise it. This symbol is lambda, a Greek letter. Pause the video and try it. Done? Okay, let's check. First, match the coefficients with the general form. The x y term is minus five x y, so h is minus five, right? No, no, careful! Remember the little two? The term is two h x y. So two h is minus five, and h is minus five by two. Same for g and f: two g is four, so g is two, and two f is minus fourteen, so f is minus seven. And the easy ones: a is one, b is six, and c is lambda. With me? Now put these into delta. First, a b c gives six lambda. Next, two f g h: two times minus seven times two times minus five by two, that is seventy, under the curly brace. Then minus forty nine for a f square, minus twenty four for b g square, and minus twenty five by four lambda for c h square. Collect: six lambda minus twenty five by four lambda is minus lambda by four, and seventy minus forty nine minus twenty four is minus three. Set delta equal to zero: lambda equals minus twelve! Puzzle solved: only lambda equals minus twelve keeps it splitting. Now the factors. The degree two part, x square minus five x y plus six y square, is x minus two y, times x minus three y. So try x minus two y plus p, times x minus three y plus q, with constants p and q to find. Compare the x terms: p plus q equals four. Compare the y terms: minus three p minus two q equals minus fourteen. Solve them together, and p is six, q is minus two. And the check: p times q is minus twelve, exactly our lambda, hence the tick mark. So the answer: x minus two y plus six, times x minus three y minus two. Absolutely perfect, and no trial and error!

8:23Quick revision: Delta = 0, half coefficients, range by D

So, quick revision. One: the expression splits into two linear factors if and only if delta, the three by three determinant, is zero. Two: h, g and f are half the coefficients of x y, x and y. Three: for the range of y, make it a quadratic in x, and put D greater than or equal to zero. Now go and try the practice problems on this page. All the best!