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 3 of 5.
Milind said:
7 years ago
(√x-1/√) -2 = (x + 1/x) - 2.
Can you tell me where this -2 comes from?
Can you tell me where this -2 comes from?
Rohit said:
9 years ago
Answer is 4 how do you get 2?
Abisheak said:
6 years ago
Where is the power 2 comes from in the starting stage? We can See that there is no power mentioned in the question.
Shaya said:
4 years ago
For what reason we are multiplying 3-2 √x / 3-2 √x? Please explain me.
PADUSHA KHAN SHAIK said:
1 year 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.
Swetha said:
1 decade ago
Can anyone please explain whole sum clearly how -2 came in the 2nd step of the sum?
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
Shahid said:
1 decade ago
Convert the expression x into (a+b)^2 form
with a=squareroot(2) and b=1
and then after that try it
with a=squareroot(2) and b=1
and then after that try it
Preethi said:
1 decade ago
Can anyone explain this sum clearly? please.
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
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