Aptitude - Square Root and Cube Root - Discussion

Discussion Forum : Square Root and Cube Root - General Questions (Q.No. 7)
7.
If x = 3 + 1 and y = 3 - 1 , then the value of (x2 + y2) is:
3 - 1 3 + 1
10
13
14
15
Answer: Option
Explanation:

x = (3 + 1) x (3 + 1) = (3 + 1)2 = 3 + 1 + 23 = 2 + 3.
(3 - 1) (3 + 1) (3 - 1) 2

y = (3 - 1) x (3 - 1) = (3 - 1)2 = 3 + 1 - 23 = 2 - 3.
(3 + 1) (3 - 1) (3 - 1) 2

x2 + y2 = (2 + 3)2 + (2 - 3)2

   = 2(4 + 3)

   = 14

Discussion:
45 comments Page 4 of 5.

Skrn said:   10 years ago
The question I got didn't have a root 3 value. Thanks.

KARTHI said:   9 years ago
How to solve this problem logically?

Deepak Gehlot said:   9 years ago
This is wrong solution.

Pranay Patil said:   9 years ago
Can anyone explain me why can't we solve x and y individually to get x^2 = 4 and y^2 = 1/4? So to get x^2 +y^2 = 17/4.

Shudipta Baruah said:   9 years ago
This is correct: 14

x = (3 + 1) x (3 + 1) = (3 + 1)2
= 3 + 1 + 23 = 2 + 3.
(3 - 1) (3 + 1) (3 - 1) 2.

y = (3 - 1) x (3 - 1) = (3 - 1)2 = 3 + 1 - 23 = 2 - 3.
(3 + 1) (3 - 1) (3 - 1) 2,
x2 + y2 = (2 + 3)2 + (2 - 3)2,
= 2(4 + 3),
= 14.

Kalesha said:   9 years ago
(2 + 3) 2 + (2 - 3) 2.
2 (4 + 3).

How can anyone explain this?

Ram said:   8 years ago
(2 + 3) 2 + (2 - 3) 2.

Explanation:
7+4root3+3+4-root3+3.
+4root3-4root3 cancel.

After that,

4+3+4+3 = 14.

Sinthu said:   8 years ago
I couldn't understand. How is it possible?
(1)

Sinku said:   8 years ago
x2+y2=(x-y)2+2xy, can we use this formula?

Anyone explain me.
(1)

Naren said:   8 years ago
Here, 2(x^2+y^2) = (x+y)^2+(x+y)^2.


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