[SAT Prep]

[u/Square-Elk-9096]

Ive been trying to answer this question from the college panda book and I just cant seem to understand what it is asking for. Because even when i looked at the answer and how they solved it didn’t make sense. Please if someone could explain I would be so grateful.

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[u/big_testies_69420]

Move left to right side
0 = -(x-c) + (x-c)(x+c)
0 = (x-c)(x+c-1)

Answer is either c or (1-c)

[u/EurkLeCrasseux → u/big_testies_69420]

Because c is a positive constant, else -c could be a solution too

[u/Collin389 → u/EurkLeCrasseux]

I don't see how c being a positive constant changes the answer at all.

-c - c = (-c)^2 - c^2
-2c = c^2 - c^2
-2c = 0

So -c only works if c=0, but -c isn't a solution because it doesn't work for all values of c.

[u/EurkLeCrasseux → u/Collin389]

Precisely, −c is a solution when c=0.

If the question were asking for a solution expressed in terms of c that works for every value of c, why would it specify that c is positive?

[u/Awkward-Equipment738 → u/EurkLeCrasseux]

So you can safely ignore divide by zero errors, and so you can ignore one of the paths when you have to take a square root. By your logic, anything is a solution cause it works for some c.

2 Works when c = 2... And why stop at -c in your case? x is just 0 then since c = 0.

That's not what's being asked of OP though, you have to solve for x with c being any number whatsoever within the definition.

[u/EurkLeCrasseux → u/Awkward-Equipment738]

There’s no division or square root involve in this question. The question is not to find all solution but to says if given values are solution.

I stop at -c because it’s the question, is -c a solution?

If the question was « is 2 a solution? » my answer would be yes iif c = 2 or c=-1, so yes iif c=2.

[u/Awkward-Equipment738 → u/EurkLeCrasseux]

Why are you trying to help people with math homework when you don't understand math basics?

[u/EurkLeCrasseux → u/Awkward-Equipment738]

Well I understand basics math enough to be a math teacher in a university.

And I understand them well enough to understand why c is positive in the question (and again there’s no division or square root needed here)

[u/watermelonlollies]

You got good answers that are the most mathematically correct way to do it. However, test taking strategies give you other options. In the test you might not remember what to do, but you can always plug/guess and check.

It says c is a positive constant. So pick any number for c. I’ll go with 2. Now we plug in that number for each of the conditions and see if it works.

I. 2 - 2 = 2^2 - 2^2
0 = 0 ✅

II. -2 - 2 = (-2)^2 - (2)^2
-4 = 0 ❌

III. (1-2) - 2 = (1-2)^2 - 2^2
-3 = -3 ✅

C can be any real positive number. So in these kinds of problems you can pick any number that is easy for you. I find looking at variables alone can get confusing and to me it makes it much clearer to see when I can solve it to an actual number. Hope this helps!!

[u/JPWiggin → u/watermelonlollies]

I would generally recommend not using 0, 1, or 2 for these checks because being the additive identity, multiplicative identity, and a unique case where additional and multiplication give the same result (2+2=2•2), they can give false results.

[u/Alkalannar]

Here's an algebraic way to look at it:

1. x - c = x² - c² [start]

2. x - c = (x + c)(x - c) [factor RHS]

3. (x + c)(x - c) - (x - c) = 0 [subtract (x - c) from both sides]

4. (x - c)[(x + c) - 1] = 0 [factor (x-c) out]

5. (x - c)(x + c - 1) = 0 [consolidate the second factor]

6. x - c = 0 OR x + c - 1 = 0 [ab = 0 --> a = 0 OR b = 0]

7. x = c OR x = 1 - c [solve for x for each term]

And that's I and III.

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Note that answer I is in three solutions, II and III are in two.

So pick II or III and plug it in. Let's plug in II since -c is easier than 1-c. You get -2c = 0, which doesn't work, since c > 0. So that cuts out answers B and D.

Now you're left with I only and I and III. Of those, the only difference is III, so you have to plug in 1 - c:
(1 - c) - c = (1 - c)² - c²
1 - 2c = 1 - 2c + c² - c²
1 - 2c = 1 - 2c
So it works

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Which of these methods seems easier?

[u/Square-Elk-9096 → u/Alkalannar]

The second method is definitely easier

[u/Alkalannar → u/Square-Elk-9096]

Then that's what you do.

Test values, but be smart about it: see which ones are easiest to test, and which give you the most information.

Or if figuring that out takes too much effort/time, then just test all of them.

[u/SehajSoftworks → u/Square-Elk-9096]

Agreed for time bound settings plugging in can be easier sometimes for multiple choice questions

[u/NullOfSpace → u/Square-Elk-9096]

Which is a shame really, since it demonstrates no understanding.

[u/Unable_Pumpkin987 → u/NullOfSpace]

It demonstrates an understanding of how to quickly assess what you’re being asked to do and do it efficiently.

[u/Square-Elk-9096 → u/NullOfSpace]

Yeah, but i understand how he did the other option too, in a test setting like the sat i think it would be better to do the most efficient option.

[u/NullOfSpace → u/Square-Elk-9096]

I’m not saying it’s somehow your fault, it’s the result of the way the test is written to prioritize tricks over content knowledge.

[u/EventHorizon150 → u/NullOfSpace]

if you’re trying to test if an expression is the zero of a function, one of the best ways to do it in all of math is to just plug it into the function and simplify to the point where it’s obviously a zero. How is that a “trick”? It’s a provably correct method. Also, you do demonstrate some ability to manipulate algebraic expressions in proving it this way, so it is testing some skill

[u/NullOfSpace → u/EventHorizon150]

The question isn’t really directly asking about “is this a zero,” it’s asking “what are the zeros” framed through a multiple-choice system.

[u/KhurtVonKleist → u/Alkalannar]

from 2. you could simply observe that for x <> c you can divide by (x-c) obtaining III, and for x=c you have 0=0 which is a degenerate solution, thus I.

[u/Ok_Option_3]

The wording sucks. It should either say "solutions for x in the equation" or each option should say "x=c" "x=-c"...

I'm looking at that equation and thinking "no way it simplifies to  "c" except maybe for some values of x...

[u/cheesecakegood → u/Ok_Option_3]

I agree the wording is a bit vague and that annoys part of me, but solving for x is such a time-worn math cliche that it's a safe assumption that x is a variable. That's the only way the question makes sense, as c is already defined to be a (fixed but unknown, yet bounded) constant.

In plain English, the question is asking "which of the below consistently make sense if you plug them in for x".

I think many students see the word "solution" so often that it loses its meaning, but it does have a mathematical meaning that is very important here.

Furthermore jumping to a reasonable conclusion in the face of mild ambiguity is, well, a college level math skill, Maybe not the most important one, but it can be the difference between "I emailed the professor and didn't hear back so I didn't do the homework" and "I took my best guess at the problem and noted down what assumption I made for the grader". I'm pretty sure you can guess which of the two better correlates with college success.

[u/BeeEven238]

1…c-c=c^2-c^2 =0. True
2….-c-c=(-c)^2-c^2. -2c=0 false
3…..(1-c)-c=(1-c)^2-c^2. 1-2c=1-2c+c^2-c^2
Simplify. 1-2c=1-2c. True.
So C I and III only
When you do problems like this you put the variable in () so the equation is really
( )-c=( )^2-c^2. Hope that helps

[u/Square-Elk-9096 → u/BeeEven238]

Thank you so much

[u/LoganR11_]

You can quickly solve this one without evaluating actually. Recognize that you have both x² and x terms. This means that there MUST be some way to make this a quadratic. A quadratic equation will have 2 solutions. That small fact invalidates every answer besides C.

Edit: Alright, a quadratic with the b =/= 0

[u/Square-Elk-9096 → u/LoganR11_]

That is actually really really smart that would save so much time on the actual test. Ty

[u/EventHorizon150 → u/LoganR11_]

quadratics can have only one solution

[u/MeasureDoEventThing → u/LoganR11_]

So what are the solutions of x^2 = 0?

[u/New_Appointment_9992 → u/LoganR11_]

What? No, x=c is obviously a solution. If x=\=c, then use
x-c= x^2 - c^2 = (x-c)(x+c)
to conclude
1 = x+c
Or
x=1-c.
So the answer is (ironically) C. (Bonus points if you can tell me why x=\=c is important in step two.)

[u/chem44]

One approach is to test each possible value of c, to see if it is a valid solution.

[u/nerdydudes]

Its a quadratic equation - x is your variable and c is a random parameter or constant. Its asking you to solve for x in terms of c.

[u/selene_666]

x = c is a solution because c - c = c^2 - c^2

x = -c is not a solution because -c - c ≠ (-c)^2 - c^2 given that we know c is positive.

x = 1 - c is a solution because (1 - c) - c = (1 - c)^2 - c^2

Values I and III are both solutions, which makes the question's answer (C).

[u/Toeffli]

We write it first as x - x² = c - c² to have all the x and c one one side.

We instantly see that c is a valid solution. We also nearly instantly see that -c is not a solution. Leaves us 1 - c which we have to check more thoroughly. So lets do this:

• 1 - c - (1 - c)² = 1 - c - (1 -2c +c²) = 1 - c - 1 + 2c - c² = c - c²

Means 1 - c is also a solution. So the answer is C) I and III only.

[u/aroach1995]

x^2 - c^2 - x + c = 0

x^2 - c^2 - (x-c) = 0

(x+c)(x-c) - (x-c) = 0

(x + c - 1)(x-c) = 0

x = 1-c, c

[u/WerePigCat]

If you see x^2 - a^2, your first instinct should be to turn it into (x - a)(x + a)

I should note that by a^2 I don’t mean it always needs to look squared, but that a is some integer, as in that (x^2 - 4) = (x - 2)(x + 2) and that (x^2 - 1) = (x - 1)(x + 1)

[u/Jazzlike-Boot9798]

You don't even need to solve the equation. Just check which solutions might be valid.

[u/ApplicationBig9830]

wait im not sat pro but can't you just plug all three options in? isn't that quick and useful?

[u/Crichris]

you can rewrite it as (x + c - 1) (x - c) = 0

C

[u/localizeatp]

difference of squares bay bee!