The Quadratic Equation and the Nature of its Roots

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Take x squared minus four x plus thirteen equals zero. In the formula, under the square root, you get minus thirty six. The square root of a negative number? So this equation has no roots at all, right? Hmm. Hold that thought. Today: the nature of the roots of a quadratic. Look, JEE loves asking what kind of roots an equation has, without solving it. So, let's begin.

0:33The quadratic equation and its formula

So, first the basics. The quadratic equation is a x squared, plus b x, plus c, equals zero, with a not equal to zero; the slashed equal sign means not equal. Unless stated otherwise, a, b and c are real, okay? Now, a number alpha, this Greek letter, is called a root if it satisfies the equation: a alpha squared, plus b alpha, plus c, equals zero. Clear? Next, the discriminant, written as capital D: b squared minus four a c. Remember this letter; it is the hero of today's lesson. And the two roots, alpha and beta, a second Greek letter, are minus b, plus or minus the square root of D, all over two a; the plus-minus sign gives one root each. You know this from school, right? But tell me, why is it true?

1:26Why the formula works: complete the square

It's simple! We complete the square. Write a x squared plus b x plus c as a, times, in big square brackets, x plus b over two a, the whole squared, minus D over four a squared. Expand it at home and check, okay? So the expression is zero exactly when x plus b over two a, whole squared, equals D over four a squared; the double arrow means if and only if. Take square roots, and that's it, there is the formula. Now look. On the left, a square. On the right, four a squared is positive. So what decides whether a real x exists?

2:03The sign of D decides the nature

Yes, exactly! It is the sign of D that decides. If D is greater than zero, we get two distinct real roots. Look at the first picture: the parabola y equals a x squared plus b x plus c, with a positive, cuts the x axis at two points. If D equals zero, the two roots are equal, both minus b over two a. In the middle picture, the parabola just touches the x axis. Got it? And if D is less than zero? The square of a real number can never be negative, and the last picture never meets the x axis. So, no roots at all, right? Just like our puzzle? No, no, careful! No real root, yes. But now we have complex numbers. There are two non-real roots, p plus or minus iota q, a conjugate pair. So our puzzle equation does have roots; we will find them in Example one. Clear?

2:58Rational coefficients: the perfect-square test

Alright, now a sharper question. Suppose a, b and c are rational numbers. When are the roots rational? Look at the formula: which part can spoil it? Yes, only the square root of D can spoil it! Everything else is rational. The curly e sign means belongs to, and the Q with a double line is the set of rational numbers. So the roots are rational exactly when D is the square of a rational number. That's all it takes! And if a root is irrational, like p plus root q, then its partner is p minus root q. Irrational roots come in pairs, okay? One more. If a, b and c are integers, that is the Z with a double line, and a equals one, then every rational root is an integer, and it divides c. Always!

3:45Topper's trick: add the coefficients

Now a trick that toppers use all the time. Put x equals one in a x squared plus b x plus c. What is left? Yes, just a plus b plus c! So if a plus b plus c is zero, then x equals one is a root, and the other root is c over a; the arrow means so. No formula needed, right? And put x equals minus one: you get a minus b plus c. So if a minus b plus c is zero, then minus one is a root. Really simple.

4:18An identity, and two traps

Now, a quick question. Can a quadratic equation have three different roots? No! Because a non-zero polynomial of degree at most two has at most two roots. So if it holds for more than two distinct values of x, then a, b and c are all zero, and it holds for every x: an identity, okay? Now two JEE traps. First, conjugate pairs. p plus or minus iota q needs real coefficients, and p plus or minus root q needs rational coefficients. With complex coefficients, the roots need not be conjugates. Don't assume it! Second, a parameter in the x squared coefficient, like m minus two, times x squared; the three dots mean the rest of the equation. Check m equals two separately: there the x squared term vanishes and the equation becomes linear. Students forget this case. Got it?

5:12Example 1: distinct, complex and equal roots from D

Now, remember our puzzle? Let's finish it. Solve three equations. Part one: two x squared minus seven x plus three equals zero. Part two, our puzzle: x squared minus four x plus thirteen equals zero. Part three: x squared minus two root three x plus three equals zero. Pause the video, and try them first. Done? Okay, let's check. Part one. D is forty nine minus twenty four: twenty five, a perfect square. So x equals seven plus or minus five, over four: three, or one half. Two distinct real roots, and rational too, because D is a perfect square! Part two, our puzzle. D is sixteen minus fifty two: minus thirty six. Negative. So now what do you say: no roots? Well, no real roots, but two non-real ones! The square root of minus thirty six is six iota. So x equals four plus or minus six iota, over two: two plus or minus three iota. So, a conjugate pair, exactly as promised. Puzzle solved! Part three. b squared is minus two root three, whole squared, that is twelve, and four a c is also twelve. So D is twelve minus twelve, zero. So the two roots are equal: x equals root three, repeated. With me?

6:44Example 2: when are the roots rational? (JEE Main 2019)

Now a harder one, straight from JEE Main twenty nineteen. How many positive integers alpha make the roots of six x squared minus eleven x plus alpha equals zero rational? Pause and try. Hint: think perfect square. Done? Okay, let's check. D is b squared minus four a c: one hundred twenty one, minus twenty four alpha. Rational coefficients, so D must be a perfect square, and so greater than or equal to zero; that is the sign with a line below. If alpha is six or more, D becomes negative. So alpha can only be one, two, three, four or five. Okay? Put them in: D is ninety seven, seventy three, forty nine, twenty five, and one. All five are positive, so all five give real roots. So the answer is five, right? No, no, careful! The question says rational, not real. That is the JEE trap. Ninety seven and seventy three are not perfect squares. But forty nine is seven squared, twenty five is five squared, and one is one squared. So alpha is three, four or five: three values!

7:55Quick revision: sign of D, perfect squares, a + b + c = 0

So, quick revision. D equals b squared minus four a c. Positive: two distinct real roots. Zero: equal roots. Negative: a non-real conjugate pair. Rational coefficients: rational roots exactly when D is the square of a rational. If a plus b plus c is zero, one is a root. Parameter in the x squared coefficient? Check the linear case. Now go and try the practice problems on this page. All the best!