Aptitude - Surds and Indices - Discussion
Discussion Forum : Surds and Indices - General Questions (Q.No. 15)
15.
| If x = 3 + 22, then the value of |  |
x | - | 1 |  |
is: |
| x |
Answer: Option
Explanation:
|
x | - | 1 | |
2 | = x + | 1 | - 2 |
| x | x |
| = (3 + 22) + | 1 | - 2 |
| (3 + 22) |
| = (3 + 22) + | 1 | x | (3 - 22) | - 2 |
| (3 + 22) | (3 - 22) |
= (3 + 22) + (3 - 22) - 2
= 4.
 |
|
x | - | 1 | |
= 2. |
| x |
Discussion:
48 comments Page 1 of 5.
Sourabh said:
3 years ago
@All.
If any root number is in the denominator so we have to a rationalization for solving that (means to remove the root number from the denominator) then in this condition we do multiply of denominator number by doing a sign change (in both nominator and denominator).
If any root number is in the denominator so we have to a rationalization for solving that (means to remove the root number from the denominator) then in this condition we do multiply of denominator number by doing a sign change (in both nominator and denominator).
(2)
Akashdeep Singh said:
8 years ago
In the last stage where we find the value of [(underoot)x - 1/(underoot)x]^2 = 4.
So if you remove the square then it will become (underoot)4 i.e. 2.
(underoot)x - 1/(underoot)x = (underoot)4.
(underoot)x - 1/(underoot)x = 2.
So if you remove the square then it will become (underoot)4 i.e. 2.
(underoot)x - 1/(underoot)x = (underoot)4.
(underoot)x - 1/(underoot)x = 2.
Aparna said:
1 decade ago
@ravi,
Your prblm is how the denominator part came 1
see
the denominator is
(3-2 2)(3+2 2)
use the formula
(a+b)(a-b)=a2(square)-b2(square)
hence
3(square)-(2(square)*2)
here suqreroot of one 2 gets removed
we get
9-8=1
Your prblm is how the denominator part came 1
see
the denominator is
(3-2 2)(3+2 2)
use the formula
(a+b)(a-b)=a2(square)-b2(square)
hence
3(square)-(2(square)*2)
here suqreroot of one 2 gets removed
we get
9-8=1
Neenu said:
1 decade ago
x= 3+ 2(root)2
=> 2+1+2(root)2.(root)1
{which is} [(root)2+1)^2
hence, (root)x=(root)2+1
1/(root)x=1/(root)2+1 => (root)2-1
So when we add them [x+1/(root)x] = (root)2+1-[(root)2-1]=>(root)2+1-(root)+1=>2
=> 2+1+2(root)2.(root)1
{which is} [(root)2+1)^2
hence, (root)x=(root)2+1
1/(root)x=1/(root)2+1 => (root)2-1
So when we add them [x+1/(root)x] = (root)2+1-[(root)2-1]=>(root)2+1-(root)+1=>2
Samarth said:
2 years ago
To explain 4th Step we are trying to remove the denominator, it's basically the formula (a+b) * (a-b) = a^2 - b^2.
If we solve a^2 - b^2, in this case, it comes to 9-8 = 1 and thus the denominator becomes 1.
If we solve a^2 - b^2, in this case, it comes to 9-8 = 1 and thus the denominator becomes 1.
(2)
PADUSHA KHAN SHAIK said:
2 years ago
Here we observe that initially, we take (√x+1/√x)=p.
Assume 'p'.
Now s.b.s.
(√x+1/√x)^2 = p^2.
Now step 4.
3 + 2√2 + 3 - 2√2 - 2 = p^2.
6-2 = p^2.
4 = p^2.
P = 2.
Assume 'p'.
Now s.b.s.
(√x+1/√x)^2 = p^2.
Now step 4.
3 + 2√2 + 3 - 2√2 - 2 = p^2.
6-2 = p^2.
4 = p^2.
P = 2.
Hima said:
1 decade ago
3+2(root)2 can be written as 1+2(1)(root2)+ (root2)^2 which is equal to (1+root2)^2 [a^2+b^2+2ab=(a+b)^2].
So
Now x=(1+root2)^2 now put this value in the question.
So
Now x=(1+root2)^2 now put this value in the question.
Harshini said:
9 years ago
In the 4th step, if you see the denominator: (a+b) (a-b) = (a^2-b^2) (using the formula).
It becomes 3^2 -(2 root 2)^2.
Which is 9 - (4 * 2),
=> 9 - 8 = 1.
It becomes 3^2 -(2 root 2)^2.
Which is 9 - (4 * 2),
=> 9 - 8 = 1.
GAURAV BHARDWAJ said:
1 decade ago
@Swetha.
If you will square that term using formulae (a-b)^2 = a^2+b^2-2ab, then you will automatically come to know how -2 came in second step.
If you will square that term using formulae (a-b)^2 = a^2+b^2-2ab, then you will automatically come to know how -2 came in second step.
Sagar sharma said:
1 decade ago
The answer should have been 1. As we put the value of x in given expression and rationalize 1/x, we get 1 as the answer.
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