Algebraic Booster #2 - Terrible Radicals
It is given that x = 1 − x 1 + x − x 1 .
Evaluate the value of x 1 5 − x 6 1 0 .
The answer is 987.
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3 solutions
DO not give such a lenthy qn.
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:P Hahaha
Yup!! I agree...this question is more of the conventional type than creative ....
So x=phi.....nice approach!
Attention! A = 1, not -1.
Your method is nice and clean. The method below is not better than your method. But just to give an alternative method. We will also use the _EQUATION _ x^2 = x + 1. You have used identity (a+b)^2 = a^2 + 2ab + b^2, below we use identity (a+b)*(a-b) = a^2 - b^2
x^2 - 1 = x ............................................ x^2 = x + 1
x^2 + 1 = x + 2
x^4 - 1 = x^2 + 2x = 3x + 1
x^4 + 1 = 3x + 1 + 2 = 3(x + 1)
x^8 - 1 = 3(3x^2 + 4x +1)= 3(3x + 3 + 4x + 1) = 3(7x + 4)
x^8 + 1 = 3(7x + 4) + 2
x^16 - 1 = 3(7x + 4) }^2 + 6(7x + 4) = 9*49x^2 + 144 + 9*56x + 144 + 42x +24
= 441x^2 +504x + 144 + 42x +24 =( 441x +441) + 546x +168
x^16 = + 1 + 987x + 609
x^16 = + 1 + 987x + 609
x^15 = 987x/x + 610/x ................. x^15 - 610/x = 987
I just asked Wolfram to solve. Good Job!
How great....
Wow. Just wow. Didn't think of that.
WTF
Can you like explain how did you got the 1 at the 7th step of the long solution you have posted! I am trying to do it using your equation but I don't know how you got the 1. I did this earlier but was stuck at the 7th step when figuring it all out on my own then I saw urs and u somehow got the 1. Please explain
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Let's say that sqrt(x) + sqrt(1+x) = 50 Imagine multiplying the left side of the equation by its conjugate. (sqrt(x) - sqrt(1-x)) If you do that, you will get x - (1+ x) = -1 It now means that (sqrt(x) + sqrt(1+x))(sqrt(x) - sqrt(1+x)) = -1 (Try writing this on paper.) We know that sqrt(x) + sqrt(1+x) = 50, then sqrt(x) - sqrt(1+x) = -1/50 We can now subtract our equations
Can somebody help me find solutions to this question?
Anagram Cracker!!
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2) geaperioitrdspawsagnhabineod
3) enaednenetorfyimrw
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Details and assumptions:
Example:
"My name is Anil" can be written in the form of group of letters as:
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Wait, wait. Why A = -1? If you plug the A into 2A = A^2 + 1, it will be -2 = 2...
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Oops. I am mistaken. A = 1. Thank you for noticing this ^-^
on solving the identity x^2 = x+1 we get to values and putting these two values in expression (x^16 - 610)/x we two answers 987 and 754. am i correct.
It was a real good question but too big!!!!!:P
NICE question but solved it diffrently. :-)
Nice solution. The way you used the Identity is really good. To arrive at the Identity, instead of using A, we could also do it as follows. Not that it is better than yours. Only an alternating method. (x - 1/x) - 2sqrt(x - 1/x) + 1 = 0 ={sqrt(x - 1/x) - 1}^2 :- x - 1/x = 1,...x^2 = x + 1.
but if i solve the eqn....x^2=x+1......then ans comes out to be 1.618......if i take this....ans is 987 but if i take 1.62 then comes out 1013......so confused..actually power 15 gives the trouble...
x = sqrt (1- (1 / x)) + sqrt (x- (1 / x))→ Call, sqrt (1- (1 / x)) = a; sqrt (x- (1 / x)) = b Notice that: a ^ 2 + (1 / sqrt (x)) ^ 2 = 1 ^ 2; b ^ 2 + (1 / sqrt (x)) ^ 2 = (sqrt (x)) ^ 2 Now draw a triangle of height (1 / sqrt (x)), and sides (a + b); 1 and sqrt (x). Note that: (a + b) * (1 / sqrt (x)) = 1 * (sqrt (x)) Thus, we have a triangle rectangle of sides 1; sqrt (x) and x. For Pythagoras, we have to: x ^ 2 = 1 ^ 2 + (sqrt (x)) ^ 2→ x = (1 + sqrt (5)) / 2. Consequently: (-1 / x) = (1-sqrt (5)) / 2 So we have to: x ^ 15-610 * (1 / x)→ 682 + 305 * sqrt (5) + 305-305 * sqrt (5)→ 987
Can somebody help me find solutions to this question?
Anagram Cracker!!
Anagrams are problems related to shuffled letters which are needed to be arranged and made into perfect meaningful sentences without repeating the letters (letters can be used only once).
Here are some anagrams which you need to crack:
1) tuteauaewribeifslh
2) geaperioitrdspawsagnhabineod
3) enaednenetorfyimrw
Remember to arrange and make a meaningful sentence (one sentence from each group of letters), not single word. If you are able to solve this anagrams please inform me the answers as well as how you found the solutions to the anagrams.
Details and assumptions:
Example:
"My name is Anil" can be written in the form of group of letters as:
meailaysmnni
If x = 1 − x 1 + x − x 1 ,
then x = x − x 1 + 1 − x 1 .
If we multiply x − x 1 + 1 − x 1 by x − x 1 − 1 − x 1 , we will get x − 1 .
So, if x − x 1 + 1 − x 1 = x , then x − x 1 − 1 − x 1 = x x − 1
x − x 1 + 1 − x 1 = x ------------------> Equation 1
x − x 1 − 1 − x 1 = x x − 1 -----> Equation 2
Equation 1 + 2 : 2 x − x 1 = x + x x − 1
2 x − x 1 = ( x − x 1 ) + 1
Let A be x − x 1
2 A = A 2 + 1
A = 1
x − x 1 = 1
x − x 1 = 1
Since x = 0 (which will make 0 1 in the question), we can multiply both sides by x.
x 2 − 1 = x
x 2 = x + 1 ------> Important Identity
So, x 1 5 − x 6 1 0
= x x 1 6 − 6 1 0
= x ( x 2 ) 8 − 6 1 0
= x ( x + 1 ) 8 − 6 1 0 (Identity)
= x ( x 2 + 2 x + 1 ) 4 − 6 1 0
= x ( 3 x + 2 ) 4 − 6 1 0 (Identity)
= x ( 9 x 2 + 1 2 x + 4 ) 2 − 6 1 0
= x ( 2 1 x + 1 3 ) 2 − 6 1 0 (Identity)
= x ( 4 4 1 x 2 + 5 4 6 x + 1 6 9 ) − 6 1 0
= x 4 4 1 x + 4 4 1 + 5 4 6 x + 1 6 9 − 6 1 0 (Identity)
= x 9 8 7 x
= 9 8 7
What a hardcore solution! :3