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 2 of 5.
Mohan said:
6 years ago
@Joe.
In the first step, it is in the form of (a-b) ^2 formula.
In the first step, it is in the form of (a-b) ^2 formula.
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?
BFC said:
7 years ago
How to solve 4th step the denominator (3+2√2) (3-2√2)? Please explain.
Mahima said:
8 years ago
4th step how to solve the denominator (3+2√2)(3-2√2)? Please explain.
Harigovind C B said:
8 years ago
if a÷b=4÷5 and b÷c=15÷16 , then (c^2-a^2)÷c^2+a^2 is equal to?
Can anyone solve this?
Can anyone solve this?
Bishal Dey said:
8 years ago
= 3+2√2 + 1/3+2√2 -2,
= 3+2√2 + 3 - 2√2 -2,
=6-2.
=4.
=√4=2.
= 3+2√2 + 3 - 2√2 -2,
=6-2.
=4.
=√4=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.
Felix said:
9 years ago
How did you obtain 1/x +x -2?
(3)
Valens said:
9 years ago
How 4 = 2 at last?
Saurav Kumar said:
9 years ago
If x+1/x =1 then (x+1)^3 + 1/(x+1)^3 = ?
Please solve it.
Please solve it.
(1)
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