Re: x=4√(x3+6x2)
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28 Jul 2021, 07:17
If presented with clear algebraic information in a Quantitative Comparison, initially look to solve algebraically.
First, raise the given equation to the 4th power to eliminate the radical resulting in x4=x3+6x2.
Next, subtract x3+6x2 from each side to set the quadratic equal to 0 in order to solve.
Then, factor x2 out of each term resulting in x2(x2−x−6) = 0.
Next, factor the interior quadratic x2−x−6 resulting in the equation x2(x−3)(x+2)=0.
Thus, it would appear that the roots are x = 0, x = 3 , and x= -2.
However, be careful to not conclude that the quantities are equal because 0 + 3 + -2 = 1.
Remember that the result of an even radical applied to a negative value is an imaginary number and thus invalid on the GRE.
So, x cannot be negative, because if it were, the result would be (4
√
(−23+6(−2)2)
=−2 which would not be an acceptable result of real arithmetic.
Subsequently, there are only two valid solutions : x = 0 and x = 3, and the sum of those valid roots will be 3 + 0 = 3, and 3 > 1.
So answer is A