Harvard University Admission Exam | Ivy League Schools Aptitude Test | Tricks You Should Know | x=?

Published: Aug 26, 2024 Duration: 00:12:20 Category: Education

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hello and welcome on how to solve this m Olympia problem 16 /x - 8 /x 2 + 4 x Cub - 2 x^ 4 + 1 / x^ 5 = 32 what is the value of x solution we are being given 16 / x - 8 / X2 + 4 / x cubed - 2 / x^ 4 + 1 / x^ 5 = 32 now here let us multiply through of this equation by x to^ 5 so we have x to^ 4 - 8 x^ 3 + 4 x² - 2 x + 1 = to 32 x^ 5 now let us bring all terms to one side of the equation so we have 32 x to^ 5 - 16 x to^ 4 + 8 x cubed - 4 x 2 + 2 x - 1 = 0 now let us express the coefficients in index form so we have 2 to^ 5 which is the same as 32 * x^ 5us 2^ 4 * x^ 4 + 2 ra^ 3 * x^ 3 - 2 2 * x 2 + 2 x - 1 equal to0 now here for every term let us apply this rule that a ra^ m * p^ m equal to a p of this ra to power M therefore we have 2X of this ra to power 5us 2 x of this to^ 4 + 2 x of this ra to^ 3 - 2 X of this to^ 2 + 2x - 1 = 0 now here let N be equal to 2x by substitution we have n l to^ 5us n to^ 4 + n to^ 3 - n to^ 2 + N - 1 = 0 now here let us change the form of this equation so we have n to^ 4 + 1 the same as 5us n l to^ 4 + n l to power 2 + 1 - n 2 + N - 1 = 0 now here let us apply this rule that a to p + C is the same as a to P * a to c so by the application of this rule then this terms changes to n to power 4 * n^ 1 - n^ 4 + n^ 2 * n^ 1 - n 2 + N - 1 equal to0 now let us factorize by grouping so let us group the first two terms the next two terms and the last two terms Now by factorization n to^ 4 is common the first two time so we have n^ 4 into N - 1 plus the next we have n² into N - 1 the third one we have N - 1 = to 0 now as you can see nus1 is common let us pull it out so we have nus1 into n^ 4 + n 2 + 1 = 0 now here let us add let us add n² - n 2 which equal to 0 on this Factor so we have n -1 into n^ 4 + n 2 + 1 + n 2 - n 2 = 0 so I have added 0 in terms of n 2 so n 2 + n 2 you get 2 n 2 so we have n -1 into n^ 4 + 2 n 2 + 1 - n 2 = 0 now next we have n - 1 into n 2 of this 2 + 2 n 2 + 1 - n 2 = 0 now here let us apply this rule on this three terms of the second Factor so apply this rule that a 2 + 2 2 a p + p² = a + P the of this squ the soal the soal pomi expansion identities so this one becomes nus one into now we have we have n 2 + 1 of d squ - n 2 equal to 0 now here let us apply the difference of two squares that a s- p^ 2 = a + P into a minus P therefore we have n -1 into n 2 + 1 + n into n 2 + 1 - n = 0 by simplifying fact and writing the standard form of quadratic we have N - 1 into n 2 + n + 1 into n 2 - n + 1 = 0 here we have three cases now here n = 1 here let us use the quadratic formula in which n = minus p plus or minus the square root of p ^ 2 - 4 a c the of this/ 2 a here A = to 1 P = to 1 c = 1 by substitution n = to- 1 + or minus the root of 1 2 - 4 * 1 * 1 1 the of this/ two so n = to -1 + or minus the square root of - 3 over 2 now here we know that -1 equals to i² so this is important here so it means that it means that and = to -1 + or minus theare root of i² * squ of 3 / 2 so which means that n = to -1 + or minus I < tk3 / 2 so in case 2 we have obtained the value of n now let us obtain the value of n in case three so here we have n equals to by use of this formula here A = to 1 p = to1 c = to 1 by substitution n = to into this for into this formula so we have 1 + or minus the square root of - 1 2 - 4 * 1 * 1 over 2 * 1 so n = to 1 + or minus the square root of now this one becomes - 3/ two by use of this complex relationship then we have n = to 1 + or minus I < tk3 over two so you can see we have obtain to values of n now recall n = 2x which means that in the first case 2x = 1 by dividing both sides by two we have x = 1 / 2 so this is the first root of our equation now next here we have by putting 2x to be equal to n so we have 2X = to -1 + orus I < tk3 / 2 so which gives us by dividing both sides by two we have x = -1 + or - I3 over 4 that is by dividing both sides by two so this is the value of x and here so we have 2X = to 1 + or minus I theun of 3 / 2 by dividing both sides by two we have x = 1 + or minus I theun of 3 / 4 now the four Solutions of our equation are now x equal to 1 / 2 comma - 1 + I < tk3 / 4 comma - 1 - I root otk 3/ 4 comma 1 + I < tk3 / 2 comma 1 minus I < tk3 this over 4 and that is it thank you for watching subscribe to my channel and turn on the notification Bell to get new updates when I uplo new videos don't forget to smash the like button share and comment below to make this video many people

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