Please, help with this equation/problem
- Thread starter mr2
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Deleted member 4993
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mr2 said:
What level of math are you doing.
If in high school, best way to solve cubic is to plot and estimate and then check.
If you know calculus - then there is a method called "Newton's method" to find real roots of a polynomial - which can be used.
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I'm guessing that the answer key gave the value that you were supposed to find, rather than the value you were supposed to "check". Is this correct? :?:
As previously mentioned, solving cubics is kind of rough, especially when you don't have calculus. :!:
The graphing method will work, of course: Go into GRAPH mode on your calculator, and, for Y1, enter [open paren] 1 [-] [var key] [close paren], etc, etc, until you have the whole expression entered (using "X" instead of "y"); then tack a "[-] 25/8" on the end. In this way, the x-intercept of your graph will be the x-value that makes (1 - x)(3 - x)(6 - x) equal to 25/8 = 3<sup>1</sup>/<sub>8</sub>, since, if they're equal, subtracting them will give you zero. :idea:
To do this by "exact" means (and I'll give this to you because I wouldn't expect this from an Algebra II student), you could do the following jiggery-pokery:
. . . . .\(\displaystyle \L (1-y)(3-y)(6-y)=\frac{25}{8}\)
. . . . .\(\displaystyle \L (3-2-y)(3-y)(3+3-y)=\frac{25}{2\times 2\times 2}\)
. . . . .\(\displaystyle \L ((3-y)-2)(3-y)((3-y)+3)=\frac{5\times 5}{2\times 2\times 2}\)
I have three factors on the left, and a cube in the denominator on the right. So I probably need everything on the left to somehow get a denominator of "2". Also, I have a square, (5)(5) = 25, on the right, and three factors on the left. If I can get the left to be in the form of (5)(5)(1), I may be on the right track. So:
. . . . .\(\displaystyle \L \left(\frac{2\left[(3-y)-2\right]}{2}\right) \left(\frac{2\left[3-y)\right]}{2}\right) \left(\frac{2\left[(3-y)+3\right]}{2}\right) = \frac{5\times 5}{2 \times 2\times 2}\)
. . . . .\(\displaystyle \L \frac{8\left[(6-2y)-4\right]\left[(6-2y)\right]\left[(6-2y)+6\right]}{2\times 2\times 2} =
\frac{5 \times 5}{2 \times 2\times 2}\)
If I had the following:
. . . . .\(\displaystyle \L (x-4)(x)(x+6)=25\)
...I might notice that, if x = -1, I would then have:
. . . . .\(\displaystyle \L (-5)(-1)(5) = (5)(5) = 25\)
Of course, I don't have the equation in terms of "x", even if we ignore the denominators. But -- if we do ignore those denominators, let "x" replace "6 - 2y", and add in 2's account for the "8" in front, I then have:
. . . . .\(\displaystyle \L \left[2(x-4)\right]\left[2(x)\right]\left[2(x+6)\right] = 25\)
Instead of needing -5, -1, and 5, I now need -5/2, -1/2, and 5/2, in order to cancel off those leading 2's and get 25 as my result. In other words:
. . . . .2(x - 4) = -5/2
. . . . .x - 4 = -5
. . . . .x = -1
But I'm letting "x" stand for "6 - 2y", so:
. . . . .6 - 2y = -1
. . . . .6 = -1 + 2y
. . . . .7 = 2y
. . . . .y = 7/2
And that's the answer they gave you. But how on earth they'd expected you to arrive at that (at your present level of study), I don't know! :shock:
Eliz.
As previously mentioned, solving cubics is kind of rough, especially when you don't have calculus. :!:
The graphing method will work, of course: Go into GRAPH mode on your calculator, and, for Y1, enter [open paren] 1 [-] [var key] [close paren], etc, etc, until you have the whole expression entered (using "X" instead of "y"); then tack a "[-] 25/8" on the end. In this way, the x-intercept of your graph will be the x-value that makes (1 - x)(3 - x)(6 - x) equal to 25/8 = 3<sup>1</sup>/<sub>8</sub>, since, if they're equal, subtracting them will give you zero. :idea:
To do this by "exact" means (and I'll give this to you because I wouldn't expect this from an Algebra II student), you could do the following jiggery-pokery:
. . . . .\(\displaystyle \L (1-y)(3-y)(6-y)=\frac{25}{8}\)
. . . . .\(\displaystyle \L (3-2-y)(3-y)(3+3-y)=\frac{25}{2\times 2\times 2}\)
. . . . .\(\displaystyle \L ((3-y)-2)(3-y)((3-y)+3)=\frac{5\times 5}{2\times 2\times 2}\)
I have three factors on the left, and a cube in the denominator on the right. So I probably need everything on the left to somehow get a denominator of "2". Also, I have a square, (5)(5) = 25, on the right, and three factors on the left. If I can get the left to be in the form of (5)(5)(1), I may be on the right track. So:
. . . . .\(\displaystyle \L \left(\frac{2\left[(3-y)-2\right]}{2}\right) \left(\frac{2\left[3-y)\right]}{2}\right) \left(\frac{2\left[(3-y)+3\right]}{2}\right) = \frac{5\times 5}{2 \times 2\times 2}\)
. . . . .\(\displaystyle \L \frac{8\left[(6-2y)-4\right]\left[(6-2y)\right]\left[(6-2y)+6\right]}{2\times 2\times 2} =
\frac{5 \times 5}{2 \times 2\times 2}\)
If I had the following:
. . . . .\(\displaystyle \L (x-4)(x)(x+6)=25\)
...I might notice that, if x = -1, I would then have:
. . . . .\(\displaystyle \L (-5)(-1)(5) = (5)(5) = 25\)
Of course, I don't have the equation in terms of "x", even if we ignore the denominators. But -- if we do ignore those denominators, let "x" replace "6 - 2y", and add in 2's account for the "8" in front, I then have:
. . . . .\(\displaystyle \L \left[2(x-4)\right]\left[2(x)\right]\left[2(x+6)\right] = 25\)
Instead of needing -5, -1, and 5, I now need -5/2, -1/2, and 5/2, in order to cancel off those leading 2's and get 25 as my result. In other words:
. . . . .2(x - 4) = -5/2
. . . . .x - 4 = -5
. . . . .x = -1
But I'm letting "x" stand for "6 - 2y", so:
. . . . .6 - 2y = -1
. . . . .6 = -1 + 2y
. . . . .7 = 2y
. . . . .y = 7/2
And that's the answer they gave you. But how on earth they'd expected you to arrive at that (at your present level of study), I don't know! :shock:
Eliz.
D
Deleted member 4993
Guest
That's clever - tedious - but clever!!
Decided not a fair question for class. Thanks for your help.