Square Root Part 2
Find Infinitely many square roots 4 0 5 8 2 1 0 + 4 0 5 8 2 1 0 + . . . + 4 0 5 8 2 1 0 + 4 0 5 8 2 1 0 + 4 0 5 8 2 1 1
The answer is 2015.
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3 solutions
You could have given a proper sequence or steps in your solution which could have made it better. Refer my solution.
The value of the number approaches 2 0 1 5 , and that appears to be the answer, but it is by all means not exactly 2 0 1 5 .
Yes, there would have to be an infinite number of square roots for the value to be exactly 2015.
Fixed it!
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In that case, the last 4 0 5 8 2 1 1 is redundant...
but last number is 4058211 then how to solve?
To solve this question, we will take an assumption.
Let: x = 4 0 5 8 2 1 0 + 4 0 5 8 2 1 0 + 4 0 5 8 2 1 0 + . . . . . .
x = 4 0 5 8 2 1 0 + x
Squaring both sides, we get:
x 2 = 4 0 5 8 2 1 0 + x
x 2 − x − 4 0 5 8 2 1 0 = 0
x 2 + 2 0 1 4 x − 2 0 1 5 x − 4 0 5 8 2 1 0 = 0
x ( x + 2 0 1 4 ) − 2 0 1 5 ( x + 2 0 1 4 ) = 0
( x + 2 0 1 4 ) ( x − 2 0 1 5 ) = 0
x = − 2 0 1 4 , 2 0 1 5
Since, x is not taken as negative.
Thus, the answer is: x = 2 0 1 5
Let the following be equal to X Then 4058210 + X = X^2 ( Squaring Both Sides) Solve for X and the value Comes out to be 2015 .