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Bunuel
The value of \(−3 + \sqrt{9−k}\) is positive for which of the following values of k ?

I. -1
II. 0
III. 1

A. None
B. I only
C. II only
D. III only
E. I and III

As per the question \(−3 + \sqrt{9−k}>0\) or \(\sqrt{9−k}>3\). Squaring both sides we get
\(9-k>9\). Hence \(k<0\). From the given values only \(-1\) satisfies this condition

Option B
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Bunuel
The value of \(−3 + \sqrt{9−k}\) is positive for which of the following values of k ?

I. -1
II. 0
III. 1

A. None
B. I only
C. II only
D. III only
E. I and III

We can analyze each Roman numeral:

I. -1

-3 + √(9 - (-1))

-3 + √10

-3 + 3.2 is positive.

II. 0

-3 + √(9 - 0)

-3 + √9

-3 + 3 is NOT positive.

III. 1

-3 + √(9 - 1)

-3 + √8

-3 + 2.8 is NOT positive.

Alternate Solution:

-3 + √(9 - k) > 0

√(9 - k) > 3

9 - k > 9

-k > 0

k < 0

Of the choices, only Roman numeral I, for which k = -1, satisfies the requirement for the value of k.

Answer: B
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I have a doubt.

In each of the cases whether we consider the value whose square root we are taking as 9, 10, or 8, can't the actual square root be either positive or negative? As a result, we cannot ascertain whether any of the values from amongst the options given (-1, 0, or +1) will for sure provide an end result that is a positive number.

Is there something I am missing here?
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B
The number is negative when k is 0 or greater than 1.
K has to be negative.

Posted from my mobile device
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