What is the value of \(x?\)
(1) \(x^2+x+10=16\)
(2) \(x=4y^4+2y^2+3\)
[spoiler]OA=C[/spoiler]
Source: Veritas Prep
(1) \(x^2+x+10=16\)
(2) \(x=4y^4+2y^2+3\)
[spoiler]OA=C[/spoiler]
Source: Veritas Prep
Let's take each statement one by one.M7MBA wrote:What is the value of \(x?\)
(1) \(x^2+x+10=16\)
(2) \(x=4y^4+2y^2+3\)
[spoiler]OA=C[/spoiler]
Source: Veritas Prep
(1) \(x^2+x+10=16\)
=> x^2 + x - 6 = 0 => x = -3 or 2. No unique value of x. Insufficient
(2) \(x=4y^4+2y^2+3\)
Since we do not have the unique value y, we can't get the unique value of x. Insufficient.
(1) and (2) together
From (2) \(x=4y^4+2y^2+3\), note that \(4y^4+2y^2+3\) is a positive value, irrespective of the value of y. Thus, from (1), one of the values of x = - 3 is ruled out. Thus x = 2. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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