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 1 of 5.
Pavan Kumar said:
2 years ago
The value of √3 is 1.732. By applying this, we will get the answer.
(5)
Bhushan said:
6 years ago
x2 + y2 = (2 + √3)2 + (2 - √3)2.
= (4+2*2*√3+3) + (4 - 2*2*√3+3),
= (4+3) + (4+3),
= 2(4+3).
= (4+2*2*√3+3) + (4 - 2*2*√3+3),
= (4+3) + (4+3),
= 2(4+3).
(5)
Jamshaid said:
3 years ago
@All.
Here, as whatever power of 1, it remains 1, simply;
(3)^1/2 +1 = (4)^1/2 = 2;
(3)^1/2 - 1 = (2)^1/2,
Thereby X = (2)^1/2 & Y = 1/(2)^1/2,
X^2 = 2 & Y^2 = 1/2.
Ans 2.5.
Here, as whatever power of 1, it remains 1, simply;
(3)^1/2 +1 = (4)^1/2 = 2;
(3)^1/2 - 1 = (2)^1/2,
Thereby X = (2)^1/2 & Y = 1/(2)^1/2,
X^2 = 2 & Y^2 = 1/2.
Ans 2.5.
(3)
Naomi Mwamba said:
4 years ago
Can someone clearly explain this in step by step?
(3)
Chandrakala Rajput said:
2 years ago
Using , (x+y)^2 = x^2 +y^2 -2xy
So, x^2 +y^2 = (x+y) ^2 - 2xy ---> (I)
(X+y) = (√3 + 1) /(√3 - 1) + (√3 - 1) /(√3 + 1).
On taking LCM,
(X+y) = {(√3 + 1)^2 +(√3 - 1)^2}/(√3 - 1)(√3 +1).
(X+y) =(3+1-2√3 +3+1-2√3) /[(√3) ^2 - 1^2].
(X+y) = 8/(3-1),
(X+y) = 8/2,
(x+y) = 4.
Now, XY = {(√3 + 1) /(√3 - 1)} * {(√3 - 1) /(√3 + 1)},
XY = 1.
Now putting values of (X+y) & XY in equation (I)
x^2 + y^2 = (x+y) ^2 - 2xy ---> (I)
x^2 + y^2 = (4) ^2 - 2,
x^2 + y^2 = 16 - 2,
x^2 + y^2 = 14.
Answer is 14.
So, x^2 +y^2 = (x+y) ^2 - 2xy ---> (I)
(X+y) = (√3 + 1) /(√3 - 1) + (√3 - 1) /(√3 + 1).
On taking LCM,
(X+y) = {(√3 + 1)^2 +(√3 - 1)^2}/(√3 - 1)(√3 +1).
(X+y) =(3+1-2√3 +3+1-2√3) /[(√3) ^2 - 1^2].
(X+y) = 8/(3-1),
(X+y) = 8/2,
(x+y) = 4.
Now, XY = {(√3 + 1) /(√3 - 1)} * {(√3 - 1) /(√3 + 1)},
XY = 1.
Now putting values of (X+y) & XY in equation (I)
x^2 + y^2 = (x+y) ^2 - 2xy ---> (I)
x^2 + y^2 = (4) ^2 - 2,
x^2 + y^2 = 16 - 2,
x^2 + y^2 = 14.
Answer is 14.
(1)
Sinku said:
8 years ago
x2+y2=(x-y)2+2xy, can we use this formula?
Anyone explain me.
Anyone explain me.
(1)
Jayshree said:
10 years ago
Hello, I just didn't understand how is 3+1+2√3/2 = 2+√3.
Please explain.
Please explain.
(1)
Sakthi said:
1 decade ago
Root 3 -> 1.732.
1.732+1=2.732.
1.732-1=0.732.
2.732/0.732=3.7322.
0.732/2.732=0.2679.
(3.7322)^2=13.929 & (0.2679)^2=0.0192.
Add 13.929+0.0192=13.9482 is near by 14.
So answer is 14.
Are you clear now friends.
1.732+1=2.732.
1.732-1=0.732.
2.732/0.732=3.7322.
0.732/2.732=0.2679.
(3.7322)^2=13.929 & (0.2679)^2=0.0192.
Add 13.929+0.0192=13.9482 is near by 14.
So answer is 14.
Are you clear now friends.
(1)
Vikas said:
1 decade ago
Could understand how {(3+2)/(3-2) }*{ (3+2)/(3+2)/(3+2)} = (3+2)^2/ (3-2), isn't it should be (3+2)^2/(3^2-1^2)? Please inform me asap.
Chotu said:
1 decade ago
(a+b)2+(a-b)2 = 2(a2+b2).
Here a = 2 =>a2 = 4; b = 3^(1/2) =>b2 = 3.
=>2(4+3).
=>2(7).
=>14.
I think this is the correct one because this is the model of rationalization. If this is not the correct answer please explain.
How would you get the answer as 2?
Here a = 2 =>a2 = 4; b = 3^(1/2) =>b2 = 3.
=>2(4+3).
=>2(7).
=>14.
I think this is the correct one because this is the model of rationalization. If this is not the correct answer please explain.
How would you get the answer as 2?
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