If x2 = 131, which of the following is closest to a potential value of

is a correct approach to this problem 11^2 = 121 and 12^2= 144. 121 is closer to 131 than 144 or so -11 is the answer??

Just know your squares for this question and that any negative number to an even exponent is always positive

144 131 121

Clearly -11 is closer

Thus
'B"

11*11 = 121 , Difference from 131 is 10
12*12 = 144 , Difference from 131 is 13
13*13 = 169 , Difference from 131 is 38

Thus, the answer must be (B)

\(x^2 = 131\)

We can see that value of \(x^2\) lies in between \(121\)- Square of \(+/- 11\) and \(144\) Square of \(+/- 12\)

If we compare in further detail, we come to know that \(131\) is more closer to \(121\) which is square of \(+/- 11\)

Hence, we choose \(- 11\) from the options.

Hence, Answer is B

We need to determine the approximate value of x and are given that x^2 = 131.

Thus, x = √131 or x = -√131.

Since √121 < √131 < √144, we see that 11 < x < 12 (if x = √131) or -12 < x < -11 (if x = -√131). Therefore, we need to see whether x is closer to -11 or 12, since both of these numbers are in the given answer choices. We see that 131 is closer to 121 than it is to 144. So:

If x = √131, x is closer to 11 than it is to 12.

If x = -√131, x is closer to -11 than it is to -12.

In other words, regardless if x is √131 or -√131, it is closer to 11 or -11 than it is to 12 or -12.

Alternate Solution:

The two perfect squares between which 131 lies are 121 and 144. We see that 121 is closer to 131 than is 144, so the perfect square of interest is 121, which we can get as either 11^2 or (-11)^2. From the given answer choices, we see that our choice must be -11.

Answer: B

\(11^2 < X^2 <12^2\)

\(121 < 131 <144\)

\(131 - 121 = 10\) ; \(144 - 131 = 13\)

So, \(11^2\) is nearest to \(131\), Answer must be (B)

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