So, I’ve recently been laid off from the Navy, and decided to use my GI benefits to go to college (hoping that when I’m done, there will be more work available). The college I chose has a placement exam for math and English. I expected to do poorly on the math portion, as I haven’t taken any math classes in about 20 years. And indeed, I am going to be taking basic college math.
One of the questions stuck with me though, and I can’t figure out what the answer was supposed to be. I suppose I could wait until April, when classes start up, but I’m impatient. The question is, what are the two numbers for X? Here goes:
X2(that is, X squared-I can’t figure out how to make superscripts))+3X-18=0
The choices given are as follows:
a) 3 and 6
b) -3 and 6
c) 3 and -6
d) 3 and 18
e) -3 and 18
f) 3 and -18
Now, based on my memory of basic algebra (again, from about 20 years ago), there should have been a different letter up there (that is, X2+3Y-18=0, but I’ll admit that I’m not 100% positive on that. My thinking was that since a negative multiplied by a negative equals a positive, then the first number is always 9. Therefore: 9+3X-18=0. But I can’t figure out what that second “x” should be. Presumably, 9+3X should equal 18 (thus, 18-18=0), but none of the options led to that.
This is a quadratic equation in standard form. Since your variable is squared, there will be two possible solutions (roots) to the equation. This particular equation can be solved by factoring the polynomial and then solving each one for x.
So given x[sup]2[/sup] + 3x - 18 = 0,
we can factor it as: ( x - 3 )( x + 6) = 0
(If you multiply that out using FOIL, you’ll see that it results in the original equation, so our factoring is correct.)
Now it’s easy to solve each part:
x - 3 = 0, so x = 3
x + 6 = 0, so x = -6
You’re missing a lot - I think you’ll find taking the math class helpful.
If you want to do this by inspection (a perfectly reasonable method), then you plug a number in for X as follows:
3[sup]2[/sup] + 3*3 - 18 = 0 check - this works, so one of the roots of the equation is 3. That means that you have to examine 6, -6, 18, and -18
6[sup]2[/sup] + 3*6 - 18 = 36 nope, 6 is not a root
(-6)[sup]2[/sup] + 3*(-6) - 18 = 0 yep, that’s the other one.
So the answer is c).
The other method is to factor the quadratic equation as friedo did, which is more general, but may be slower on a multiple choice timed test.
BTW, a great resource for reviewing a lot of basic (and not that basic) math topics is Khan Academy. They have a video on solving quadratic equations by factoring here.
I think **friedo **gave an excellent explanation and this is just how I would have explained it to someone who was in the middle of this class.
In your case, since your a little rusty, let me clear up one problem with the way you’re trying to think through this.
I think your confusion here is that it is true that x can have two values, but only one at a time. You cannot use one value of x to determine x[sup]2[/sup] and another value of x to determine 3x. You have to solve the entire equation consistently using one value of x. The reason that there are two solutions for x is that x[sup]2[/sup] can have a positive x and a negative x that both give the same term.
I have nothing to add except that you can do [sup]superscript[/sup] and [sub]subscript[/sub] by using the [sup] and [sub] tags. Just don’t forget to close them!
Note to the OP. While factoring works in this case, it does not always work. So pay attention in class when they cover the quadratic formula. That will always work.
To go a bit further in depth, both solutions can be negative and both can be positive. That’s because the two solutions for x for the general quadratic equation ax[sup]2[/sup] + bx +c = 0 are:
x = (-b + sqr(b[sup]2[/sup] - 4ac))/2a
x = (-b - sqr(b[sup]2[/sup] - 4ac))/2a
So the difference between the two solutions is the + or - in front of the square root bit.
Factoring does always work, it’s just that the factoring might be rather tricky and produce non-integer solutions (since, after all, sometimes the correct solutions are non-integers)…
Thanks a lot, guys. I’m definitely going to be looking at the Khan Academy site, as there seem to be a lot of helpful videos there.
ENugent’s equation helped me see what I was trying to do (which I’m guessing is considered the “scenic route”), but I see that my assumption (that it was 9+3X-18=0) was wrong.
CWG, when you say
do you mean in the same way that while 2+2=4, so does 1+3? In other words, was the question asking, not, "What is the one answer to the problem, but what are all possible (at least within the choices given) answers to the problem? And why is this tied to the variable being squared (like freido said)?
And will this be clearer once I start taking the classes?
If x[sup]2[/sup] = 4, the correct answer to “find the value of x” would have to be “2 or -2.”
The problem you posted is just a slightly more complicated version of “find the value of x.” Squaring a variable means you need to account for the fact that a square root can be either positive or negative, and then the problem you posted complicates it by subtracting a number (called a constant, in your case 18) and adding what’s called a cross term (in your case, 3x).
Take a quick look at the ‘roots’ (i.e. the answer to your question), the ‘alternate forms’ (which are the ones people have shown above, with the same expression in a different form), and the ‘plots’.
The plots bit basically shows what you’d see on a graph, where the y axis for any given x value is equivalent to x[sup]2[/sup] + 3x - 18.
The answer for x[sup]2[/sup] + 3x - 18 = 0 is looking for what are called the ‘roots’ of this function, which are the values of x where y is 0 (i.e. where the graph crosses the horizontal line). So if y is 0, there are two different values you might have been able to put in as x to end up with y being 0.
The example posted by Ace309 is a simpler version of this same type of thing. If you are looking for x where x[sup]2[/sup]=4, we know that x[sup]2[/sup] must be equal to 4, so 2 or -2 are both perfectly legitimate values for x to ensure that x[sup]2[/sup] = 4. This is the same as asking for what x is when you know that x[sup]2[/sup] - 4 = 0
First, in case you start asking lots of algebra questions, another commonly accepted way to write x[sup]2[/sup] is just using the caret (^) character instead of superscript, like this: x^2. Nearly as easy to read, and much quicker to write on a computer.
Yes, exactly.
Well, for now, just accept that equations can have as many real solutions as the highest exponent (but might have fewer). So if the biggest exponent is x^2, there can be 0, 1 or 2 real solutions. I think asking ‘why?’ right now is just going to open up cans of worms that you probably would not have wriggling around you right now.
Don’t know, but if it’s not clearer, you can always come back here for smartest, hippest people on the planet, plus a few total dipsticks.
I think there will always be the same number of solutions as the greatest exponent. Some solutions just might be equal to each other, and/or ‘imaginary’ (xi).
May I just say that I like your reasoning? It’s wrong, of course, but it’s the kind of mistake made by someone who is unfamiliar with the conventions of algebra—not the kind of mistake made by someone who has trouble thinking mathematically.
As noted, the problem was to “fill in the blanks” of the equation [sup]2[/sup] + 3 - 18 = 0, but you weren’t supposed to find one number for each blank, you were supposed to find all possible numbers (of which there are two) that work in both blanks.
Just so this doesn’t confuse you, it’s not true that the roots must be one positive and one negative. x[sup]2[/sup]-4x+3 = 0, for example, has roots of 1 and 3.
?
Solve the following: x[sup]2[/sup] = 0
The only solution is 0, unless you want to get silly and say “The solutions are 0 and 0,” in which case you might as well say “The solutions are 0 and 0 and 0 and 0 and 0 and 0…”
I can’t think of an equation right now, but a 3rd-power equation might have roots of -1, -1, and 5. So there are three roots, but two of them are the same. Is ±0 valid? I don’t know, since zero isn’t signed. (Someone will answer, I’m sure.)
**Thudlow Boink **has the most direct response to this particular question:
You assumed the first x must be 3 or -3, because of the possible answers, and tried to figure out what the second x was. But you have to use the same value of x all the way through.
