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Given, x is a real number

Quantity A:

$$(x-2)(x-4)(x+5)$$

Quantity B:

$$(x-5)(x^2+5x+5)+68$$

The first thing I did was to expand Quantity B and after combining like terms I get

$$x^3-20x+43$$

Since $43$ is prime I can't (I don't think) factor that so I decide to subtract that from Quantity A, but first I expand quantity A and, after combining like terms, I get

$$x^3-x^2-22x+40$$

so after subtracting quantity B from quantity A I get

$$x^3-x^2-22x+40-x^3+20x-43$$ $$-x^2-2x-3$$

Immediately I can see that the discriminate of that is less than zero so the roots will be imaginary which violates the real number constraint that was given so I select "can not be determined". However, the answer key acknowledges the negative discriminate but then says that because the roots are imaginary that quantity B is bigger. Could someone help me understand how they came up with that?

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  • $\begingroup$ I'd like to point out that $-x^2-2x-3 = -(x^2+2x+1+2) = -[(x+1)^2+2]$ which is always $\le -2$. $\endgroup$ – steven gregory May 19 '18 at 15:01
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You were right up until the part about the negative discriminant. This correctly implies that there are no real roots, but it is this which ensures that there is a definite answer. A real polynomial which has no roots is either always positive or always negative, and since $-x^2-2x-3$ is clearly negative at $x=0$, it is always negative, and hence Quantity B is bigger than Quantity A for any real value of $x$.


If you had instead found the quadratic to have positive discriminant, then there would be two real roots, and the quadratic would be positive between the roots and negative elsewhere (or vice versa), so that Quantity B could have been greater or less than Quantity A depending on $x$.

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  • $\begingroup$ This is a great answer missing only one explanation, which clarifies the exact point where the OP slipped up: @Dean MacGregor, the very very simple point to realize is that the expressions shown do not have "= 0" at the end; that is, they are expressions (quantities), not equations. (Expressions cannot have roots; only equations can have roots.) $\endgroup$ – Wildcard Dec 18 '16 at 11:12
  • $\begingroup$ @Wildcard To be totally precise about that, I would have written the OP's last equation as $B-A=-x^2-2x-3$. It is not being equated to zero. I will disagree with your last sentence; expressions (or more precisely functions) are what have roots, since a root of $f$ is a solution to the equation $f(x)=0$. An equation has solutions, but the "roots" terminology suggests equating to zero, which you can't do with an arbitrary equation. $\endgroup$ – Mario Carneiro Dec 18 '16 at 21:41
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Since you do not have a graphing calculator available on the GRE, one way to get quick properties of a parabola is to convert it to vertex form, by completing the square. First, you have to make the $x^2$ coefficient $1$ by factoring: $$A-B = -(x^2 + 2x + 3)$$ and then add and subtract $\left(\dfrac{2}{2}\right)^2 = 1$: $$-(x^2+2x+1-1+3)$$ Notice that we can factor: $$-(x+1)^2-(-1+3) = -(x+1)^2-2 = -[x-(-1)]^2+(-2)\text{.}$$ Look at this website for example: it implies that $(-1, -2)$ is the vertex of this parabola. Furthermore, since we have a $-$ sign on the squared term, that means that the parabola is pointing down from $(-1, -2)$. So, this means that $A - B$ is always negative, hence $A - B < 0$, or $B > A$.

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    $\begingroup$ Both completing the square and graphing overkill. All that's needed is the discriminant. $\endgroup$ – symplectomorphic Dec 18 '16 at 5:17
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First, you don't mean "determinant"; you mean "discriminant."

Second, the fact that the discriminant of that quadratic is negative means that it has no real roots. It follows that your quadratic function is always positive or always negative (in other words, it has the same sign for all values of $x$). But clearly it will always be negative.

Now, after you subtracted quantity B from quantity A, you changed the problem to the equivalent question of how $-x^2-2x-3$ compares to 0. Given that the quadratic is negative (thus less than zero) no matter the value of $x$, the answer is that quantity B is larger.

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If you graph the result, the parabola "points down", and since the roots are not real, it does not intersect the x-axis, so it's always negative. That means that Quantity A - Quantity B is always negative, so Quantity B must be greater.

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