Common Roots

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Here are two quadratics: x squared plus k x plus twelve equals zero, and x squared minus seven x plus ten equals zero. They have a common root, one root that satisfies both equations. Find k, okay? Most students find one value of k and move on. But tell me, how many values of k are there? Hold that thought; we will settle it in the first example. And remember, common root questions come up in JEE again and again. So, let's begin.

0:38What a common root gives us

So, take two quadratics in general. The first is a one x squared, plus b one x, plus c one, equals zero, and the second is a two x squared, plus b two x, plus c two, equals zero. The small one and two written below the letters only tell us which equation a coefficient belongs to, okay? Now suppose alpha, this Greek letter, is a common root. That means alpha satisfies both equations. So, a one alpha squared plus b one alpha plus c one is zero, and a two alpha squared plus b two alpha plus c two is zero. Clear? Now a smart idea. Pretend these are two ordinary linear equations. Then what are the two unknowns? Yes, exactly! Alpha squared is one unknown, and alpha is the other. So we can simply cross multiply, okay? And cross multiplication gives this. Alpha squared over b one c two minus b two c one, equals alpha over c one a two minus c two a one, equals one over a one b two minus a two b one. Lots of letters, but it is only cross multiplication. Got it?

1:43The condition for a common root

Now read this chain in pairs. The last two parts give alpha: c one a two minus c two a one, over a one b two minus a two b one. Of course, this bottom must not be zero. The crossed equal sign means not equal, okay? And the first and the last parts give alpha squared: b one c two minus b two c one, over the same bottom. Alright, so now we have alpha, and separately, alpha squared. But alpha squared is just alpha into alpha, right? So how do we connect these two results? Alpha squared equals alpha, whole squared, absolutely right! This double arrow means implies. Square the first result, set it equal to the second, clear the bottoms, and we get the condition: c one a two minus c two a one, whole squared, equals b one c two minus b two c one, into a one b two minus a two b one. And it works both ways, clear? One more question. When do the two quadratics share both roots, not just one? That's it, when one equation is just a multiple of the other! That is, a one over a two equals b one over b two equals c one over c two. Same ratio for all three coefficients. Simple, right?

2:56Topper's trick: kill the x² term

But in practice, there is a faster way. Here is the topper's trick. Multiply one equation by a suitable number, so that both x squared terms become the same, and then subtract one equation from the other. The x squared terms cancel, and only a linear equation is left. Solve it, and you get the common root. Then put it back into either equation to find the unknown. Quick and clean! We will use this in example two.

3:24The JEE trap: non-real roots

Now a trap that JEE has used. Take a quadratic with real coefficients, whose roots are not real. Such roots come in a pair: alpha, and alpha bar, the conjugate of alpha. That is the little bar on top, okay? Now this quadratic shares the root alpha with another quadratic, which also has real coefficients. One root is shared, so only one root is common, right? No, no, careful! The second quadratic also has real coefficients. So if alpha is its root, then alpha bar is its root too. The conjugate comes along with it, you see? So both roots are common, and the two equations are proportional, all three coefficients in the same ratio. So in a common root question, first check for non-real roots. Got it?

4:10Example 1: find k so the two quadratics share a root

Now, remember our puzzle from the start? Let's finish it. Find k, if x squared plus k x plus twelve equals zero, and x squared minus seven x plus ten equals zero, have a common root. Pause the video, and try it first. Done? Okay, let's check. The first equation has the unknown k in it, but the second one has only numbers. So, which equation can we solve right away? The second one, yes! x squared minus seven x plus ten factorises as x minus two, into x minus five. So its roots are two and five, and the common root must be one of these, okay? Try two first. Put x equals two in the first equation: four, plus two k, plus twelve, equals zero. So two k is minus sixteen, and k is minus eight. Minus eight. So we write k equals minus eight, and move on, right? No, no! Five can also be the common root. Put x equals five: twenty five, plus five k, plus twelve, equals zero. So five k is minus thirty seven, and k is minus thirty seven by five. So k is minus eight, or minus thirty seven by five. Two answers! See? This is the second value hiding in our opening puzzle. Always test every root, alright?

5:34Example 2: kill the x squared term, find beta gamma/lambda

Now a harder one, from JEE Main. Lambda, beta and gamma are just more Greek letters, and lambda is not zero. Alpha and beta are the roots of x squared minus x plus two lambda equals zero. Alpha and gamma are the roots of three x squared minus ten x plus twenty seven lambda equals zero. Find beta gamma over lambda. Pause and try. Hint: kill the x squared term, the topper's trick. Done? Okay, let's check. Alpha is in both equations, so alpha is the common root. The first has one alpha squared, the second has three. So, to cancel them, by what number should we multiply the first equation? By three, of course! Three into alpha squared minus alpha plus two lambda, minus, three alpha squared minus ten alpha plus twenty seven lambda, equals zero. The alpha squared terms cancel. Collect the rest, and seven alpha equals twenty one lambda. So alpha equals three lambda. Just a linear equation, as promised! Now substitute back. Put alpha equals three lambda in the first equation: nine lambda squared, minus three lambda, plus two lambda, equals zero. Lambda is not zero, so we can divide by it, and lambda is one ninth. Then alpha, three times lambda, is one third. Now the other roots, from the sum of roots. In the first equation, alpha plus beta is one, so beta is two thirds. In the second, alpha plus gamma is ten by three, so gamma is three. With me? Almost there. Beta gamma is two thirds into three, which is two. And lambda is one ninth. So, beta gamma over lambda is how much? Careful with the division! Eighteen, that's right! Two divided by one ninth means two into nine, so beta gamma over lambda is eighteen. Some students write two ninths here in a hurry. Don't do that, okay? See? One linear equation did all the work. Well done!

7:46Quick revision: kill x squared, the condition, non-real trap

So, quick revision. For a common root, kill the x squared term, solve the linear equation, and substitute back. The condition: this bracket, whole squared, equals the product of these two brackets. And for real quadratics with non-real roots, if one root is common, then both roots are common, so all three coefficients are in the same ratio. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!