PLEASE HELP; I don't know whether I should x-box factor these or just solve and if I should just solve how do I solve it?????? (1.) x^2+2x-24a=0 (2.) x^2-14x+33=0
hmmi dont think u x-box it...but im not so sure
-24a?
im sorry i can't help :(
With a square you generally factor one way or another to find the possible zeros, but with -24a in the one, it would really change things...
its just -24 I missed typed it but because it =0 IDK what to do
You want numbers that multiply to become the last number, but add to become the center number. Their sign gives a clue. For example, if the last number is positive, then it can only be made by multoplying two positives or two negatives. If it is negative, then one is positive and the other is negative.
please draw a pic of what I should do on the first two problems using x-box and the chart for x-box PLEASE @e.mccormick drawings help me alot
If the y are the same sign, then the middle term tells you if it is two + or - numbers. Two - add to be -, and two + add to be +. If you have a + and - that are being multiplied, then you need to look at if the middle is + or - to see what is the larger. If it is +, then the + factor is larger than the - factor. If it is -, then the - is the larger (further from 0) factor.
That is the math behind it that they use in the x-box.
but what number would go where on the x and on the chart I need @ least 1 or 2 examples to do this sheet. I'm still losttttt?????
Because there is no number in the a spot, well, it is just 1, these should be simpler. \(ax^2+bx+c\)
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you aren't really helping with the question?? I understand all that you've said but is till don't know how to do the problem??
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ok, now what do I do in the chart? it has 4 boxes??
I do not generally use the X box method, so I had to look it up to refresh on it. I have not used that one in a while!
on 1.) wouldn't the 24 be -24 and 2 2.) -14 on the x part and 33
Finally... Yes, -24
And I see from your PM that you got it... OK. So, lets look at the same sort of thing on the second (and hope the site does not go to loading on me again!) |dw:1386897507574:dw| Because ac is + and b is -, it must be two - factors.
I already have that part idk what the other two numbers are though??
Always start with the factor pairs of ac. What numbers multiply to 33? It is really a limited list there.
So? Have any pairs of number that multiply to 33 and are both negative?
The easy pair is -1 and -33. But those do not add to -14, so they can't be the answer.
I'm going to take a different approach, partly because I'm not familiar with the "x box" approach you're using. Starting with your x^2 + 2x - 24 = 0 (and assuming that the "-24a" was a typo), I'd re-write this as 1x^2 + 2x - 24 = 0, and identify the first and last coefficients as 1 and -24. Make a list of possible factors of the PRODUCT of these two coefficients (which is -24): -24 = -24 * 1, or 24 * (-1), or -24 = -12 * 2, or 12 * (-2), or . . or -24 = -6 * 4, or 6 * (-4). Note how the last pair of factors COMBINE to +2, which is the coefficient of the middle term of the quadratic that we're trying to factor. Rewrite this quadratic as x^2 + 6x - 4x - 24 = 0, noting that that 6x - 4x simplifies to 2x, as desired. Now factor this last equation by grouping: x^2 + 6x = x(x + 6), and -4x - 24 = -4(x+6), and note the common factor (x+6)(x-4). Factoring out this common factor, x^2 + 6x - 4x - 24 = 0 = (x+6)(x-4) = 0. Thus, x = -6 and x = 4. Check these results in the original equation, x^2 + 2x - 24 = 0. I've seen others using the "box method" of factoring quadratics and know that it's good. The method shown above works great for me. Take your pick.
Regarding the second equation, x^2 - 14x +33 = 0, or 1x^2 - 14x + 33 = 0, multiplying the first and last coefficients together results in 33, which has the factors 1 * 33 or (better) 3 * 11, or (best) (-3) * (-11). Note how 3 and 11 add up to 14 and -3 and -11 to -14. Complete this factoring.
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