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 |
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.
Shyam said:
1 decade ago
Rationalizing the denominator. Makes it very easy and find xy.
Anjani said:
1 decade ago
We can do help from the equation (x2+y2)=(x+y)2-2xy;
Priya said:
1 decade ago
How did we get 2(4+3) ???
SAMART BOY said:
1 decade ago
(2 + 3)2 + (2 - 3)2
= 2(4 + 3) ?
Can you explain?
= 2(4 + 3) ?
Can you explain?
Vipin Kumar said:
1 decade ago
How came x = 2+3 and y = 2-3?
Bhavesh Joshi said:
1 decade ago
Is there any shortcut method to solve ?
Mahesh said:
1 decade ago
Hello priya and jaggu ,we can get 2(4+3) as follows:
(a+b)2+(a-b)2=2(a+b).................(1)
Let us prove this
Take LHS of eq(1)
a2+b2+2ab+a2+b2-2ab
Here +2ab and -2ab get cancelled and the remaining can be written as
2a2+2b2
2(a2+b2)
So here a=2 and b=rt3
2(4+3)
(a+b)2+(a-b)2=2(a+b).................(1)
Let us prove this
Take LHS of eq(1)
a2+b2+2ab+a2+b2-2ab
Here +2ab and -2ab get cancelled and the remaining can be written as
2a2+2b2
2(a2+b2)
So here a=2 and b=rt3
2(4+3)
Mohit said:
1 decade ago
Can any buddy help me to understand this problem solution in a simple and detail way.
Rizy said:
1 decade ago
Solve by the normal method using formula,
(a+b)2 = a2 + 2ab + b2.
(a-b)2 = a2 - 2ab + b2.
By using this formula you will get,
8 + 6 = 14.
(a+b)2 = a2 + 2ab + b2.
(a-b)2 = a2 - 2ab + b2.
By using this formula you will get,
8 + 6 = 14.
Abhinav said:
1 decade ago
What if we keep the value of root 3 and proceed? the answer will be 10.
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