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:
71 comments Page 6 of 8.

Krishna said:   1 decade ago
Sorry you are wrong
4*a*a - 4*a + 1 can be written as
4*a*a - 2*a -2*a +1

Taking 2*a common
2*a(2*a-1) -(2*a - 1)

Hence (2*a-1)(2*a-1) but it is given as (1-2*a)(1-2*a) is completely wrong how can he get that ????

If im wrong please explain me ???

Mitz said:   1 decade ago
In this case ans. Can also proceed with (2a - 1)^2 ??? Plz clarify !!

Sarath said:   1 decade ago
But I get (a-1/2) using (-b + -sqrt (b2 - 4ac) /2a).

Sundar said:   1 decade ago
Formula: a2 + b2 - 2ab = (a - b)2

Manju said:   1 decade ago
After solving we got (2a-1)^2 but according to answer we have to take like this:

(2a-1)^2 = [-1(1-2a)]^2 (take -1 common from (2a-1))

= (-1)^2(1-2a)^2 (here -1*-1=1)

= 1(1-2a)^2

= (1-2a)^2.

Iegarry@gmail.com said:   1 decade ago
Let me explain this. When you solve the the equation under the square root, you can have either (2a - 1) or (1 - 2a). According to the question a = 0.1039.

Hence,
2a-1 = 2* 0.1039 -1 which would be negative. You cannot have a negative number inside the square root though.

Hence we have 1 - 2a.

Abhilash said:   1 decade ago
According to the solution we have to follow! that is called reasoning!

Rathi said:   1 decade ago
@Souji.

(1-2a)2 = (2a-1)2 = 4a2+1-4a.

So no problem.

Souji said:   1 decade ago
@Latha I have a doubt in your explanation.

You said a=2a, b=1.

While substitute you did(1-2a).

But according to your explanation it is wrong.

Actually it comes(2a-1).

Latha said:   1 decade ago
a2-2ab+b2 = 4a2-4a+1.
(a-b)2 it is in the form.

Where a=(2a), b=1.
So, (1-2a)2.

4a2-4a+1^2 = 1-2a^2.
1-2a+3a = a+1.

0.1039+1 = 1.1039.


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