If (x^2+1)y=3, which of the following is not a possible value for y?
A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2
The OA is E.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
If (x^2+1)y=3, which of the following is not...
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We have y(x^2 + 1) = 3LUANDATO wrote:If (x^2+1)y=3, which of the following is not a possible value for y?
A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2
The OA is E.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
=> x^2 + 1 = 3/y => x^2 = 3/y - 1 => x = √(3/y - 1)
x = √[(3 - y)/y]
Since x is a real number, y must be less than equal to 3, else (3 - y) would be a negative number and x would then be an imaginary number.
Thus, any value greater than 3 is not qualified for y. The correct answer is 7/2 = 3.5.
The correct answer: E
Hope this helps!
-Jay
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We are given that (x^2 + 1)y = 3.BTGmoderatorLU wrote:If (x^2+1)y=3, which of the following is not a possible value for y?
A. 3/2
B. 2
C. 8/3
D. 3
E. 7/2
The OA is E.
I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Since x^2 is always nonnegative, we see that x^2 + 1 must always be greater than or equal to 1. Thus, in order for the product of a number greater than or equal to 1 and y to be 3, y has to be less than or equal to 3. Thus, y cannot equal 7/2.
Answer: E
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