Sumukh19 wrote:
The mean of three numbers \(2x\), \(3x\), \(x^2\) is \(12\), where \(x<0\). What is the range of the three numbers?
A. 108
B. 144
C. 176
D. 228
E. 256
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Given: The mean of three numbers \(2x\), \(3x\), \(x^2\) is \(12\)
So we can write: \(\frac{2x + 3x + x^2}{3} = 12\)
Simplify the numerator: \(\frac{x^2 + 5x}{3} = 12\)
Multiply both sides of the equation by \(3\) to get: \(x^2 + 5x = 36\)
Subtract \(36\) from both sides: \(x^2 + 5x - 36 = 0\)
Factor: \((x + 9)(x - 4) = 0\)
So, either \(x = -9\) or \(x = 4\)
Since we are told \(x<0\), we can be certain that \(x = -9\)
If \(x = -9\), then the values of the three numbers are as follows:
\(2x = 2(-9) = -18\)
\(3x = 3(-9) = -27\)
\(x^2 = (-9)^2 = 81\)
Range = (largest number) - (smallest number)
\(= 81 - (-27)\)
\(= 108\)
Answer: A