Grammar Fascist

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UsernameGrammar Fascist
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Quadratic equations are characterized by this equation: ax 2 + bx + c = 0 where a, b, and c are all real numbers not equaling zero. These are called coefficients, quadratic coefficients, to be precise. In order to have a quadratic, you must have something that looks like that equation, or something that can be arranged to look like that equation. If you just have one variable with no powers, then you have a standard algebraic equation. E.g. 2x 2 + 4x + 3 = 0 <--quadratic; 2x + 4 = 3x - 1 <-- standard algebraic. Hope that answers your first question--if not, please restate your question again so I can better understand and answer it.

Now, all the equations you listed were standard algebraic equations. I'm not sure how you learned how to solve them, but I always learned to get the numbers with the variables attached to them(x, t, y, etc.) on one side of the equation, and the normal numbers on another. Let's take that first question that you got wrong as an example:

3x+4= 4x+3

First, let's subtract 3x from both sides to get the "x's" on one side. 3x's on the left cancel and we're left with: 4 = x + 3 (Subtract 3x from 4x)

From there, we subtract three from both sides, leaving us with 1 = x...and there you have it! 1 = x. If you plug 1 back in for x, you'll see that it lines up.

We'll do one more, and then I'll let you do the rest by yourself, using this method.

-2(x-3)=-12

First, we want to eliminate any parentheses from the equation. That's pretty much a general rule for algebraic equations. So, multiplying -2 throughout the parentheses, we get: -2x + 6 = -12

Now, let's subtract 6 from both sides to get the number with x attached to it by itself. We end up with: -2x = -18

From here, we divide both sides by - 2 to get x = ___ Working it out, we find that x = 9 (-18/-2 = 9 [two negatives cancel, leaving us with a positive])

Thus, x = 9. Once again, you will find that the answer works with 9 as x.

So, if you identify your equation as a standard algebraic equation, then go ahead and try to get all the numbers with variables attached on one side and the numbers without variables on the other. From there, it's pretty easy to solve for your lone variable. Hope that helps! I suggest you try this out a couple of times to get the hang of it. If it doesn't work, just repost and I"ll try to help you further!

Best of luck in you algebraic pursuits!

Warm Regards,
Grammar Fascist
Feb 12, 2014