Stanford Math Tournament | Give It a Try!

Added: 2026-05-29

[00:00:02]Welcome to InfyGyan, friends. Today in this video, we'll be solving one very interesting question from SMT 2019 for the real values of x. So, let's get it started by writing our denominators cannot be equal to zero. So, x cannot be five, cannot be six, cannot be seven, cannot be eight.

[00:00:25]Now, in denominators, we have five, six, seven, and eight.

[00:00:34]5² - 1 is 24.

[00:00:38]6² - 1 is 35. 7² - 1 is 48.

[00:00:45]And 8² - 1 is 63.

[00:00:50]So, we can write our numerator with respect to the square minus one, which is available in the denominator.

[00:00:58]So, I can write x² - 10x + 25 - 1 and in denominator x - 5.

[00:01:13]Plus x² - 12x + 36 - 1 over x - 6 equal to x² - 14x + 49 - 1 over x - 7 plus x² - 16x + 64 - 1 over x - 8.

[00:01:56]Now, we'll pick three terms from numerators.

[00:02:02]And we'll write their perfect squares.

[00:02:06]So, x squared minus 10x plus 25.

[00:02:09]This is whole square of x minus five.

[00:02:13]Then we have minus one.

[00:02:15]In denominator, we'll write x minus five.

[00:02:18]Plus x minus six whole square minus one over x minus six equal to x minus seven whole square minus one over x minus seven plus x minus eight whole square minus one over x minus eight.

[00:02:48]Now, we will split numerators with respect to denominators.

[00:02:53]So, x minus five whole square over x minus five, we'll write x minus five.

[00:02:59]Then we have minus one over x minus five.

[00:03:05]Plus x minus six minus one over x minus six.

[00:03:12]Equal to x minus seven minus one over x minus seven plus x minus eight minus one over x minus eight.

[00:03:29]Now, x plus x is 2x. We'll subtract 2x from both the sides.

[00:03:36]So, we will cancel x or remove x.

[00:03:40]Now, minus five minus six is minus 11.

[00:03:43]We will write here minus 11 minus one over x minus five minus one over x minus six will be equal to minus seven minus eight minus 15 minus one over x minus seven minus one over x minus eight Now we will take minus 15 to the LHS.

[00:04:11]And we will take these two variable terms to the RHS. So one side we have constant 15 minus 11 on other side we will write one over x minus five plus one over x minus six minus one over x minus seven minus one over x minus eight.

[00:04:28]Let's write here.

[00:04:30]15 minus 11 equal to one over x minus five plus one over x minus six minus one over x minus seven minus one over x minus eight.

[00:04:51]Now we will write first term one over x minus five and last term of RHS.

[00:04:58]one over x minus eight Then we'll write one over x minus six minus one over x minus seven will be equal to 15 minus 11 four.

[00:05:14]Now we'll be taking LCM.

[00:05:17]So I can write here x minus eight minus of x minus five which will be minus x plus five over x minus five times x minus eight plus x minus seven minus of x minus six which is minus x plus six over x minus six times x minus seven equal to four.

[00:05:55]Plus x minus x will get over.

[00:06:00]We will write minus eight plus five minus three over.

[00:06:05]Let's multiply these two brackets. We'll get x squared minus five x minus eight x minus 13 x minus five times minus eight plus 40.

[00:06:17]Then we will get minus seven plus six minus one. So, I'm writing here minus one.

[00:06:24]Multiplication of these two brackets will be x squared minus six x minus seven x minus 13 x minus six times minus seven plus 42.

[00:06:36]equal to four.

[00:06:39]Now, here we have x squared minus three 13 x plus 40 and here x squared minus 13 x plus 42.

[00:06:48]So, let us assume the average equal to substitution. We will write here x squared minus 13 x plus 41.

[00:07:02]Suppose this is equal to u.

[00:07:05]So, here we can write u minus one and this denominator will become u plus one.

[00:07:15]So, we can write our equation after the substitution minus three over u minus one minus one over u plus one equal to four.

[00:07:27]We'll write minus three over u minus one minus one over u plus one equal to four.

[00:07:38]Now, we'll multiply our equation by negative one.

[00:07:42]So, let me write here whole equation times -1.

[00:07:49]We will get 3/ u - 1 + 1/ u + 1 = -4.

[00:08:01]Now, we will take LCM.

[00:08:04]So, I will write here 3 * u + 1 + u - 1 over u - 1 * u + 1 u ^ 2 - 1 = -4.

[00:08:21]Let's simplify numerator 3u + 3 + u -1 over u ^ 2 - 1 = -4.

[00:08:37]Now, we will cross multiply 3u + u will be 4u.

[00:08:42]3 - 1 will be 2.

[00:08:45]= -4 * u ^ 2 - 1. So, we can write 4 - 4u ^ 2.

[00:08:54]Now, we will write all the terms to one side, but before that we will divide both sides by 2.

[00:09:01]So, we can write 2u + 1 = 2 - 2u ^ 2.

[00:09:10]Let's write all the terms to LHS.

[00:09:13]We will get 2u ^ 2 + 2u 1 - 2 is -1 = 0.

[00:09:24]So, we have one quadratic equation.

[00:09:26]Let me write here 2u ^ 2 + 2u -1 = 0.

[00:09:36]Now, we will use completing the square method.

[00:09:40]So, we can write here 2u squared plus 2u in LHS and one to the RHS.

[00:09:49]Minus one will be plus one in right-hand side. Now, we'll multiply our equation by two.

[00:09:55]So, I will get 4u squared plus 4u equal to 1 * 2 will be two after multiplying this equation by two.

[00:10:10]Now, we have to add one to both sides.

[00:10:13]So, we will write here plus one. We'll write here plus one.

[00:10:18]Now, these three terms will give us one perfect square of 2u plus one.

[00:10:25]And in RHS, we have three.

[00:10:29]Now, we'll take a square root from both the sides.

[00:10:32]So, 2u plus one will be equal to plus minus the square root of three.

[00:10:40]So, we can get here 2u value just by subtracting one both sides minus one plus minus the square root of three.

[00:10:52]Now, we will divide both sides by two.

[00:10:55]So, we will get here two in the denominator. In LHS, we will get u. So, this is nothing but u.

[00:11:03]And u was our substitution.

[00:11:06]x squared minus 13x plus 41 So, we will write here x squared minus 13x plus 41 equal to minus one plus minus the square root of three divided by Again, we are going to use completing the square method.

[00:11:30]So, I will subtract 41 first from both the sides.

[00:11:35]So, plus and minus 41 will get cancelled out.

[00:11:39]In RHS, we are going to get minus one minus 82 minus 83 plus minus square root of three over two.

[00:11:52]And in LHS, we have x square minus 13 x.

[00:11:59]Let me write this value over here.

[00:12:02]x square minus 13 x will be in the LHS and in RHS minus 83, let me give some space and write minus 83 plus minus square root three over two.

[00:12:21]Now, we have coefficient of x in LHS.

[00:12:26]minus 13 Half of minus 13 is minus 13 over two.

[00:12:31]The square of minus 13 over two 169 over four. So, we have to add 169 over four to both sides.

[00:12:44]Now, LHS will provide us one perfect square of x minus 13 over two.

[00:12:52]And in RHS, we will simplify.

[00:12:56]We will write here this is four.

[00:12:59]So, we will multiply here with two.

[00:13:03]So, we will multiply here in the numerator also with two. We will get minus 166 plus minus two square root three plus 169 with common denominator four.

[00:13:21]169 minus 166 will be three.

[00:13:26]So, three plus minus two is square root three over four.

[00:13:34]Now, I will write both the values here.

[00:13:37]X minus 13 over two whole is square equal to three plus two is square root three over four and three minus two is square root three over four.

[00:13:55]Now, root three is 1.732.

[00:13:58]Double of that would be greater than three. So, this quantity would be always negative.

[00:14:06]Here, we have a square. The square is negative. We are going to get complex solutions. So, this value we can reject as we are looking for real X.

[00:14:19]Now, we will accept first value as it is positive. We'll take a square root both sides. So, X minus 13 over two will be equal to plus minus is square root of three plus two root three over two.

[00:14:40]Square root four is two.

[00:14:42]Now, we have to add 13 over two both sides and get our final answer. Let me write here 13 over two plus minus is square root of three plus two is square root three over two. Our denominator is same. We will get X equal to 13 plus minus is square root of three plus two root three over two.

[00:15:16]So, we have two real values of X which will satisfy our rational equation. I hope friends you will like this video.

[00:15:24]Thank I hope friends you will like this video. Thank you so very much for watching. Do not forget to like, share, and subscribe. Bye-bye till next video.

[00:15:33]Good luck. Take care. Goodbye.