Aptitude - Square Root and Cube Root - Discussion

Discussion Forum : Square Root and Cube Root - General Questions (Q.No. 6)
6.

If a = 0.1039, then the value of 4a2 - 4a + 1 + 3a is:

0.1039
0.2078
1.1039
2.1039
Answer: Option
Explanation:

4a2 - 4a + 1 + 3a = (1)2 + (2a)2 - 2 x 1 x 2a + 3a

   = (1 - 2a)2 + 3a

   = (1 - 2a) + 3a

   = (1 + a)

   = (1 + 0.1039)

   = 1.1039

Discussion:
70 comments Page 1 of 7.

Pallavi said:   3 years ago
Any anyone explain this clearly.
(2)

Uyog said:   3 years ago
Guys..this will make you understand,
Its 4a^2 * 4a+1... i.e a * 2ab+b^2.where a=2a...(4a^2) and b=1...
So, we got formula of (a-b)^2..removing root..i.e (a-b).

We get,
(2a-1)+3.
(2 * 0.1039- 1) + 3(0.1039).
(0.2078-1) + 0.3117.
0.7922 + 0.31 17.
1.1039.
(2)

Mukesh Chauhan said:   4 years ago
Two answers are possible here.

One is: 2a-1 and 1-2a because the square of both numbers are the same.

Anyone, please clear this.
(13)

Siri said:   4 years ago
I am not understanding this whole solution, can someone explain to me in an easy manner?

Khushi said:   4 years ago
Why w (1-2a) ^2, whekat ey not (2a-1) ^2?

Please tell me.
(2)

Chinnapa reddy said:   5 years ago
I have one doubt about the solution to the problem why should we take 1-2*a only, there is another chance to take equation as 2*a-1 please tell me a solution.
(1)

Dinesh Choudhury said:   5 years ago
2a<1 so use (1-2a)^2.

But (2a-1)^2 is a imaginary number.
(3)

Varsha jadhav said:   6 years ago
@All.

We can take positive or negative square root that is (2a-1) or -(2a-1) => (1-2a) but according to the given option we need a positive value if we consider the value (2a-1) then the answer will be negative.

=> (2a-1)+3a.
=> 5a-1,
=>5*0.1039-1,
=>0.5195-1,
=>-0.4855.

but if we put the value (1-2a) then the calculation will be;
=>(1-2a)+3a.
=>1+a,
=>1+0.1039,
=>1.1039.
(3)

Surya said:   7 years ago
Why not 2a-1?

How can 1-2a?

Sintu Kumar said:   7 years ago
@Krishna.

But, my factor is (2a-1)(2a-1).

Than 2a-1+3a,
=5a-1,
=5*0.1039-1.
=0.5195-1=-0.4805.
(1)


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