Equations Reducible to Quadratics

Transcript

0:00Introduction

Hello everyone! Alright, one quick question before we start. Here is an equation: x squared, minus five mod x, plus six, equals zero. Mod x is the modulus of x, okay? It has an x squared, so it looks like a quadratic. So how many real roots? Two, right? Hmm, not so fast. Today you will count them yourself. And look, JEE loves such equations: modulus, powers, square roots, logs. One smart substitution turns each of them into a simple quadratic. So, let's begin.

0:36One substitution: modulus and powers

So, the first type: a x squared, plus b mod x, plus c, equals zero. The two straight bars around x are the modulus, read as mod x, okay? This mod x spoils the quadratic, so put t equals mod x. Mod x is never negative, so t is greater than or equal to zero; that is this sign on screen. But there is still an x squared in the equation. How do we write x squared in terms of t? Yes, exactly! x squared is the same as mod x, whole squared. Squaring removes the sign anyway, right? So x squared is t squared, and this double arrow means, so. We get a t squared, plus b t, plus c, equals zero: a pure quadratic in t. Clear? Same idea for powers: a p to the power two x, plus b p to the power x, plus c, equals zero. Put t equals p to the x. Then p to the two x is just t squared: again a quadratic in t. And p to the x is always positive, so t is greater than zero. Now suppose your quadratic in t gives a negative value of t. What will you do with it? Reject it, absolutely! Mod x is never negative, and p to the x is always positive, so for powers even t equals zero is rejected. Look, this is a favourite JEE trap: students solve for t, then keep the negative value too. Don't do that, okay?

2:02Square roots: square, then check

Next, square roots. Root A, plus or minus root B, equals C, where A and B are expressions in x. This tick-shaped sign is the square root, and the plus over minus means plus or minus, okay? The method: keep one root alone on one side, then square both sides. But after squaring, you solve and get some answers. Will all of them satisfy the original equation? Not always! Squaring can create extra roots, called extraneous roots. They satisfy the squared equation, but not the original one. Careful, they look like answers, but they are fake! So always put every answer back into the original equation. Or keep track of the sign: if root A equals B, then B must be greater than or equal to zero, because a square root is never negative. Got it?

2:53Topper's trick: IIT-JEE 1978

Now a real question, from IIT, JEE nineteen seventy eight: root of x plus one, minus root of x minus one, equals one. Toppers have a fast way for this. Look at the left side. What can we multiply it with, so that the roots disappear? Exactly, its partner, with a plus sign in the middle! Minus bracket into plus bracket: the roots get squared, so we get x plus one, minus the bracket x minus one, which is just two. No roots left! Clear? But the minus bracket is equal to one, right? So the plus bracket must be two. Now add the two equations: two root of x plus one equals three. So root of x plus one is three by two. Square it, subtract one, and x is five by four. We squared at the end, so we check: three by two, minus one by two, is one. This tick mark means it matches. That's it, done in four lines!

3:51More types, same idea

Now, a few more types, same idea. First, the reciprocal equation: a x to the four, plus b x cubed, plus c x squared, plus b x, plus a, equals zero. The coefficients read the same from both ends, okay? Look, the powers go four, three, two, one, zero. Two is right in the middle. So what should we divide the whole equation by? Right, by x squared! Divide everything by x squared, and put t equals x plus one over x. Now it is a quadratic in t. But for real x, mod t is at least two, so reject any t with mod t less than two. Next, four brackets multiplied together equal k, where a plus d equals b plus c. Pair x plus a with x plus d, and x plus b with x plus c. Both products begin with x squared, plus a plus d into x. Call that t, and that's it: a quadratic. For logs, put t equals log of x to the base a; the small a below is the base. Careful: the base must be positive and not equal to one, that is the crossed equals sign, and the argument must be positive. Got it?

5:04Example 1: solve x² - 5|x| + 6 = 0, product of the roots

So, remember our equation from the start? Let's finish it. Solve x squared, minus five mod x, plus six, equals zero, and find the product of its roots. Pause the video, and try it first. Done? Okay, let's check. Put t equals mod x, so t squared, minus five t, plus six, equals zero. That gives t equals two or three. Both are positive, so both are allowed. So the roots are two and three, and the product is six. Right? No, no, careful! t is mod x, not x. Mod x equals two gives x equals plus or minus two, and mod x equals three gives x equals plus or minus three. Four real roots! See, that answers our hook. Now the product. Two into minus two is minus four; three into minus three is minus nine. Minus four into minus nine: thirty six. Not six, thirty six! Clear?

6:05Example 2: sum of roots of x² + |2x - 3| - 4 = 0

Now a harder one, from JEE Main twenty fourteen. Find the sum of the roots of x squared, plus mod of two x minus three, minus four, equals zero. Pause and try. Hint: the mod has a whole expression inside, so split into cases. Done? Okay, let's check. Inside the mod, two x minus three is zero at x equals three by two: our critical point, okay? On one side the mod opens as it is; on the other, with a minus sign. Case one: x greater than or equal to three by two. The mod opens as it is: x squared, plus two x, minus seven, equals zero. The quadratic formula gives x equals minus one, plus or minus two root two. So both of these are roots, right? Or should we check something first? Of course, check the case condition! Minus one plus two root two is about one point eight three; the wavy equals sign means approximately. That is more than three by two: keep it. Minus one minus two root two is negative: reject it! Case two: x less than three by two. The mod opens with a minus sign: x squared, minus two x, minus one, equals zero. So x equals one, plus or minus root two. Now you do the same check. Which of these two survives the condition, x less than three by two? Correct, only one minus root two! One plus root two is more than three by two: reject it. One minus root two is negative, so less than three by two: keep it. So the sum is minus one plus two root two, plus one minus root two. The ones cancel, and we get root two. Adding all four values is the trap. Always check each root against its own case, okay?

7:59Quick revision: substitute t, reject bad t, check roots

So, quick revision. Spot the repeating block, call it t, and solve the quadratic in t. Reject impossible values of t: mod x is never negative, p to the x is always positive, and mod of x plus one over x is at least two. And after squaring or splitting into cases, check every answer. Now go and try the practice problems on this page. All the best!