Chapter Summary: Mixed Solved Problems on Quadratic Equations

Transcript

0:00Introduction

Hello everyone! Here is one quick question before we start. x, minus two over x minus one, equals one, minus two over x minus one. How many roots does this equation have? Easy, isn't it? The same fraction sits on both sides, so cancel it, and x equals one. One root, done. Right? Hmm. Hold on to that answer; in a few minutes we will see if it survives. Today is our chapter summary on quadratic equations: three tools for mixed JEE problems, and two solved past-paper problems with their traps. So, let's begin.

0:37Tool 1: domain first, then solve

So, tool number one, and we start with our puzzle. Here it is on screen. Look, the same fraction, two over x minus one, sits on both sides with a minus sign, okay? Cancel the same fraction from both sides, and we are left with x equals one. Now one small habit that toppers have: put x equals one back into the original equation. What goes wrong? Yes, exactly! At x equals one, x minus one becomes zero, so two over x minus one means dividing by zero. Undefined! So x equals one was never in the domain. So x equals one is not allowed, and the equation has no root at all. Not one root: zero roots! See, the cancelling was fine. The mistake was forgetting the domain. That is tool one: domain first, then solve. Clear? And the same care everywhere. For logs, the base must be positive and not equal to one, and the argument positive; this equals sign with a slash through it means not equal. And after squaring, extra roots can appear, so check every answer. We need the log rule in Example one.

1:46Tool 2: Vieta's sum, product and D

Tool number two. Take the quadratic a x square plus b x plus c equals zero, with real coefficients a, b, c, and a not equal to zero, and call its roots alpha and beta, these two Greek letters, okay? Quick check. Without solving anything, what is the sum of the roots, and what is the product? Yes, absolutely! Alpha plus beta equals minus b over a, and alpha beta equals c over a. These are Vieta's relations: the sum and the product come straight from the coefficients. Next, the discriminant: D equals b square minus four a c. Its sign tells you the kind of roots, again without solving. Look at the figure. D greater than zero: the parabola cuts the x axis twice, so two distinct real roots. D equals zero: it just touches the axis, so equal roots. D less than zero: it never meets the axis, so there is no real root; the roots are non-real. Keep that third picture in mind. Tool three is built on it.

2:49Tool 3, topper's trick: positive for every x

Now tool three, the topper's trick of this chapter. Take the expression a x square plus b x plus c, with a positive. So the parabola opens upwards, like all three in the figure. Here is my question. When is this expression positive for every real x, no matter which x you put in? Yes, exactly when D is less than zero! See the third picture: the whole parabola floats above the x axis. This double-headed arrow means if and only if, so it works both ways. Now, what about D equals zero? Careful. The parabola touches the axis, so at one point the value is zero, and zero is not positive, right? So D equals zero fails. And D greater than zero? The graph dips below the axis between the roots. So the condition is strict: D less than zero. Example two uses exactly this. Got it?

3:43Example 1: product of roots is 7, find k for real roots

Now, Example one, from IITJEE nineteen eighty four. The product of the roots of x square minus three k x, plus two e to the power two l n k, minus one, equals zero, is seven. For which k are the roots real? Here e is the base of natural logs, and the l n you see is the natural log. Pause the video and try it. Hint: tool one first. Done? Okay, let's check. Tool one: domain first. l n k needs k greater than zero, okay? Then e to the power two l n k is simply k square. That's it, the scary term is gone. Now tool two. The product of the roots, alpha beta, is c over a. Here a is one, so the product is just two k square minus one, and that equals seven. This arrow means implies: k square equals four. k square is four, so k equals plus or minus two. Two answers, right? Wait, careful! Remember tool one: k must be positive. So k equals minus two is rejected, and only k equals two survives. Last, are the roots really real? Check the discriminant. Here b is minus three k, so b square is nine k square. And four a c? Here a is one, and c is two k square minus one, which we just found is seven. So four a c is four times one times seven; this dot means times. At k equals two: thirty six minus twenty eight, that is eight, greater than zero. So the roots are real, and the answer is k equals two. Clear?

5:20Example 2: unequal real roots for every real b, find a

Now a harder one, from IITJEE two thousand three. Find all values of a for which x square, plus a minus b times x, plus one minus a minus b, equals zero, has unequal real roots for every real b. Pause the video and try it. Hint: tool three. Done? Okay, let's check. Unequal real roots means D greater than zero, strictly, okay? Here D is a minus b, whole square, minus four times, one minus a minus b. And it must be positive for every real b. Now expand it and collect the terms. Look carefully: D is a quadratic in which letter? Yes, in b! Because b is the one that runs over every real number. D equals b square, plus four minus two a times b, plus a square plus four a minus four. Careful: here b is the variable, and a is just a fixed number. Now tool three. D must be positive for every b, so the discriminant of this quadratic in b should be less than or equal to zero, right? No, no, careful! Not less than or equal. If that discriminant were zero, D would touch zero at one value of b, and there the roots would be equal. But we need unequal roots for every b. The coefficient of b square is one, positive, so we need four minus two a, whole square, minus four times, a square plus four a minus four, strictly less than zero. Expand. Four minus two a, whole square, is sixteen minus sixteen a plus four a square. And four times the bracket is four a square plus sixteen a minus sixteen. The four a squares cancel, right? Sixteen plus sixteen is thirty two, and minus sixteen a minus sixteen a is minus thirty two a. So thirty two minus thirty two a is less than zero, which gives a greater than one. Discriminant of the discriminant: once you see it, you never forget it!

7:15Quick revision: domain first, Vieta and the discriminant

So, quick revision. One: domain first. No zero denominators, log arguments positive, and check every answer, just like our puzzle, which had no root at all. Two: the sum of the roots is minus b over a, the product is c over a, and unequal real roots means D greater than zero. Three: with a positive, the quadratic is positive for every x exactly when D is less than zero, strictly. Now go and try the practice problems on this page. All the best!