quadratic equation question

[paul.dellorusso]

there is a question on the math section of a ManhattanPrep practice test that gives the following:

z^2-3z-4=0 --> so z=4 or -1

it then asks to compare A) z^2 to B) 12+2z

If you plug in 4 or -1 for both then in both cases Quantity B is bigger than Quantity A. What I don't understand is why you can only compare them with the same values, as opposed to using 4 to plug into A and -1 to plug into B.

If you use any possible value of z then you get cases where A is bigger and cases where B is bigger. I was inclined to answer this question saying you cant determine which one because of that, but the correct answer given was B only.

What is the rationale for only comparing with the same values when there are multiple possible values?

[prashantnigam16]

When a variable has a set of values, it means that it can assume any of those values, but only one at a time, i.e. at any instance it has a unique value which stays uniform throughout the calculation. So, as long as the same "face variable" is used in an equation/inequation, the numeric value chosen will stay same throughout the system of variables. However, if the problem statement had something like "z1, z2 are the two roots of the given equation" as its part, and have asked to compare (z1)^2 and 12+2*(z2) (or vice versa), then your approach would have been correct. This is because the face variables are different and could individually pick up any value from the available set of values, like z1=4, z2=-1 or any other permutation. This concept protracts for polynomials of any degree (including 2, quadratic in this case). Plus, look at your solution, z=4 OR -1, this explicitly means one of the values but one at a time.

HTH

[tommywallach]

That explanation is exactly right. I totally understand your confusion, but once you define x as a certain thing, it must always be that thing. If it didn't then even the equation wouldn't make sense. Imagine you solve the original equation:

z^2-3z-4=0 --> so z=4 or -1

Now, couldn't you conceivably plug your solutions BACK into the original equation, and use 4 for the first z (the one in z^2) and -1 for the second z (the one in 3z)? Of course not, because the equation would not longer be "true." So even though the equation solves you to two possible solutions, from that standpoint, z must be the same at any given moment.

Hope that helps!

-t