Location of Roots

Transcript

0:00Introduction

Hello everyone! Let's start with one small challenge. Take x squared, plus m x, minus two, equals zero. For which values of m does the number one lie between its two roots? Most students reach for the quadratic formula, and with m inside a square root, it gets messy, right? Here is the secret: you never need the roots at all. By the end of this video, you will answer it in one line. And remember, JEE asks where the roots lie again and again. So, let's begin.

0:33The key condition: k between the roots

So, let f of x be a x squared, plus b x, plus c, and capital D, b squared minus four a c, is our old friend, the discriminant. Today's method in one line: draw the parabola, and read off the conditions. Clear? Look at the middle picture. The parabola opens upward, so a is positive, and the dashed line at x equals k sits between the roots. At k, is the curve above the x axis, or below? Below, yes! Between its roots, an upward parabola dips under the axis, so f of k is less than zero. And if a is negative, everything flips: f of k is greater than zero. Now here is the clever part. In both cases, a and f of k have opposite signs. So their product, a times f of k, is negative. One condition for both cases. Neat, isn't it? And here is the topper's trick. This condition, a f of k less than zero, is the single most useful condition in this topic. It says k lies strictly between the roots, and automatically guarantees real, distinct roots. So D greater than zero comes free, because the curve must cross the axis on both sides of k, right?

1:54Both roots greater than k

Next, the first picture: both roots greater than k, right of the dashed line. Three conditions. First, real roots: D greater than or equal to zero. That sign with a line below means greater than or equal to, okay? Second, k must not sit between the roots. Between means a f of k less than zero, so here we want a times f of k greater than zero. Clear? Now think carefully. Real roots, and k not between them. Does that put both roots on the right of k? No! Both roots could be on the left of k too, right? You see, a f of k greater than zero cannot tell left from right. So, condition three: the x coordinate of the vertex, minus b over two a, which sits midway between the roots. If it is greater than k, both roots are on the right. And for both roots less than k? Same first two conditions; just flip the vertex: minus b over two a less than k. Never forget the vertex condition!

2:59Two numbers k1 and k2: roots inside or outside

Now two numbers, k one and k two, with k one the smaller; the little one and two are subscripts. Third picture: both roots inside k one, k two. The round brackets mean the end points are not included. Can you guess the conditions? Yes! Real roots: D greater than or equal to zero. And neither end between the roots: a f of k one, and a f of k two, both greater than zero. Plus the vertex between the dashed lines: k one, less than minus b over two a, less than k two. Exactly! And the reverse, k one and k two both between the roots? Use our trick twice: a f of k one less than zero, and a f of k two less than zero.

3:46The JEE trap: exactly one root in (k1, k2)

Now, exactly one root between k one and k two. Then the curve is above the axis at one end, below at the other. So f of k one, times f of k two, is less than zero. Simple, right? But here is a common trap in JEE questions. What if, for some value of the parameter, a root sits exactly at k one? Then f of k one is zero, and the product is zero, not negative. The test misses that case, though the other root may still be inside! So always test the boundary values of the parameter by solving the equation there.

4:23Signs of the roots

Got it? Last idea: the signs of the roots. Both roots positive? Then D greater than or equal to zero, the sum of the roots positive, and the product positive. Two positive numbers always give a positive sum and a positive product. With me? And roots of opposite signs? Just one condition: the product is negative, that is, c over a less than zero. But there is no D condition here! Why? Alright, look: f of zero is just c. So c over a negative means a c is negative, that is, a times f of zero is negative. Our trick with k equals zero! Zero lies between the roots, so real roots come free. That's it!

5:07Example 1: for which m does 1 lie between the roots

Now, remember our hook? Let's finish it. For which m does one lie between the roots of x squared, plus m x, minus two, equals zero? Pause the video, and try it first. Done? Okay, let's check. Here k is one, so we want a times f of one less than zero. And a, the coefficient of x squared, is one, positive. So we need f of one less than zero. Put x equals one: one, plus m, minus two, that is m minus one. So m minus one is less than zero, and the double arrow means implies: m is less than one. One line, no quadratic formula. Hook solved! But wait. We never checked the discriminant. Should we? No need! The trick says a f of k less than zero guarantees real, distinct roots by itself. So no separate D check. Nice and clean.

6:06Example 2: find m so that both roots exceed 3

Now a harder one. Find m so that both roots of x squared, minus six m x, plus nine m squared, minus two m, plus two, equals zero, exceed three. Pause and try. Hint: three conditions. Done? Okay, let's check. Condition one, real roots. Here b is minus six m, so we use D by four, D divided by four, same sign as D, okay? With a equal to one, it is three m, the whole squared, minus the constant term: two m minus two. Greater than or equal to zero, so m is greater than or equal to one. Condition two: a f of three greater than zero, with a equal to one. Put x equals three and simplify: nine m squared, minus twenty m, plus eleven, which factorises as nine m minus eleven, times m minus one. Greater than zero. Both brackets must have the same sign: so m less than one, or m greater than eleven ninths. Many students stop here. Is that the answer? No, no, careful! That was only condition two. For m less than one, condition one fails: no real roots at all! And condition three, the vertex: minus b over two a here is three m. Three m greater than three gives m greater than one. All three together: m at least one; m less than one or greater than eleven ninths; m greater than one. Common part: m greater than eleven ninths. Each condition has a job. Skip one, and the answer goes wrong. Okay?

7:48Quick revision: af(k), D and the vertex, then test the ends

So, quick revision. k between the roots: just a f of k less than zero, nothing else. Both roots on one side of k: D greater than or equal to zero, a f of k greater than zero, and the vertex. Exactly one root between k one and k two: f of k one times f of k two less than zero, then test the ends. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!