Is Choice A Correct for Comparing GRE Math Quantities?

[MartinV279]

I was going through some GRE Quantitative Practice Tests, and found this question. Maybe I'm not understanding the question right (not a native speaker, and have never solved math problems in English), but choice B doesn't seem correct to me.
Solutions for x (Quantity A) are 6 and -1, and for y(Quantity B) are -3 and -3. Meaning, -3<6 and -3<-1.
Shouldn't A be correct?
Also, there's an explanation at the bottom which is correct in general but can't seem to understand how it proves the correct choice.

 

[jedishrfu]

Yeah, I think you're right. The answer makes no sense. Also the explanation seems to be geared toward a problem of quantity A=c value and quantity B=k value.

Perhaps you can send it to someone at the GRE center for clarification.

 

[Mark44]

The answer given doesn't match the question, as far as I can see, so my guess is that there is a significant typo in the question. In the answer shown, they talk about factors (x + 2) and (x + 3).
The equation in x that they give is equivalent to x² - 5x -6 = 0, or (x - 6)(x + 1) = 0, so x = 6 or x = -1.
The equation in y is equivalent to (y + 3)² = 0, so y = -3.

For the two equations, both values of x are larger than the single y value.

 

[aikismos]

Unless I miscalculated, if $y = -3$ then $x^2 - 5x + 18 = 0$ which gives $x$ the value ${5 \pm i\sqrt{47}} \over {2}$. Since reals and complexes cannot be compared, I would say D! (I think there's a mistake in the printing.)

 

[Mark44]

Unless I miscalculated, if $y = -3$ then $x^2 - 5x + 18 = 0$

I don't see how you got your last equation. The two equations are independent of one another, so how does a value for y result in $x^2 - 5x + 18 = 0$?

which gives $x$ the value ${5 \pm i\sqrt{47}} \over {2}$. Since reals and complexes cannot be compared, I would say D! (I think there's a mistake in the printing.)

 

[aikismos]

I don't see how you got your last equation. The two equations are independent of one another, so how does a value for y result in $x^2 - 5x + 18 = 0$?

<EDIT> Ignore this post. Misread the scanned problem. </EDIT>

Since you have $E_1 = E_2 = E_3$, you merely solve $E_2 = E_3$ first, and then substitute your value of $y$ back into $E_1 = E_2$.

$E_2 = E_3 : -y^2 + 3y = 9y + 9$
$-y^2 + 3y = 9y + 9 \rightarrow (y + 3)^2 = 0 \rightarrow y = -3$

$E_1 = E_2 : x^2 - 5x = -y^2 + 3y$
$x^2 - 5x = -y^2 + 3y \rightarrow x^2 - 5x = -1 \cdot (-3)^2 + 3 \cdot (-3) \rightarrow x^2 - 5x = -18 \rightarrow x^2 - 5x + 18 = 0$

 

[aikismos]

Unless I miscalculated, if $y = -3$ then $x^2 - 5x + 18 = 0$ which gives $x$ the value ${5 \pm i\sqrt{47}} \over {2}$. Since reals and complexes cannot be compared, I would say D! (I think there's a mistake in the printing.)

Oh, wait, one could take the magnitude of both numbers!

 

[Mark44]

Since you have $E_1 = E_2 = E_3$, you merely solve $E_2 = E_3$ first, and then substitute your value of $y$ back into $E_1 = E_2$.

$E_2 = E_3 : -y^2 + 3y = 9y + 9$
$-y^2 + 3y = 9y + 9 \rightarrow (y + 3)^2 = 0 \rightarrow y = -3$

$E_1 = E_2 : x^2 - 5x = -y^2 + 3y$
$x^2 - 5x = -y^2 + 3y \rightarrow x^2 - 5x = -1 \cdot (-3)^2 + 3 \cdot (-3) \rightarrow x^2 - 5x = -18 \rightarrow x^2 - 5x + 18 = 0$

I don't think the intent was to find a simultaneous solution; i.e., a solution (x, y). My take is that the intent was merely to compare the solutions to the two equations, in which case answer A would be the correct response.

 

[aikismos]

I don't think the intent was to find a simultaneous solution; i.e., a solution (x, y). My take is that the intent was merely to compare the solutions to the two equations, in which case answer A would be the correct response.

Sorry. Small screen display. I saw the problem wrong. :D