Numerical Methods

[thamsa]

The difference between two numbers is 3. If the larger number is added to its square root, and the smaller number is added to its square, the product of the two differences equals-66.33. We want to determine the two numbers to within 10^-3.

(2.1) The two numbers, say x and y, can be determined by:

(a) simply using a non-programmable calculator.
(b) solving the equation (2x + 3)(x + sqrt[x + 3])(x² + x) = 0, and then deducing the value of y from the equation y = x - 3.
(c) solving the equation (2x + 3)(x + sqrt[x + 3])(x² + x) = 0, and then deducing the value of y from the equation y = x + 3.
(d) solving the equation x² - sqrt[x + 3] - 25.11 = 0, and then deducing the value of y from the equation y = x + 3.
(e) none of the above.

 

[Ishuda]

The difference between two numbers is 3. If the larger number is added to its square root, and the smaller number is added to its square, the product of the two differences equals-66.33. We want to determine the two numbers to within 10^-3.

(2.1) The two numbers, say x and y, can be determined by:

(a) simply using a non-programmable calculator.
(b) solving the equation \(\displaystyle \, (2x\, +\, 3)\left(x\, +\, \sqrt{x\, +\, 3\,}\right)\left(x^2\, +\, x\right)\, =\, 0,\,\) and then deducing the value of y from the equation y = x - 3.
(c) solving the equation \(\displaystyle \, (2x\, +\, 3)\left(x\, +\, \sqrt{x\, +\, 3\,}\right)\left(x^2\, +\, x\right)\, =\, 0,\,\) and then deducing the value of y from the equation y = x + 3.
(d) solving the equation \(\displaystyle \, x^2\, -\, \sqrt{x\, +\, 3\,}\, -\, 25.11\, =\, 0,\,\) and then deducing the value of y from the equation y = x + 3.
(e) none of the above.

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

As a hint to get you started, let x be the smaller of the two numbers then
y = x + 3.
Let a be the larger plus its square root
Let b be the smaller plus its square.
What equation results from (y-x) (a-b) = 66.33? What equation results from (y-x) (b-a) = 66.33? Is there more than one answer?

 

[stapel]

The difference between two numbers is 3. If the larger number is added to its square root, and the smaller number is added to its square, the product of the two differences equals-66.33.

What is the significance of the hyphen before the "66.33"? Is that supposed to be a "minus" sign? Are we supposed to take "66.33" as exact, or are we supposed to assume that the value is actually \(\displaystyle \, 66\, \dfrac{1}{3}\,\)?

We want to determine the two numbers to within 10^-3.

What methods or algorithms have they given you for this? What are they wanting you to do with the "to within" specification?

(2.1) The two numbers, say x and y, can be determined by:

(a) simply using a non-programmable calculator.
(b) solving the equation \(\displaystyle \, (2x\, +\, 3)\left(x\, +\, \sqrt{x\, +\, 3\,}\right)\left(x^2\, +\, x\right)\, =\, 0,\,\) and then deducing the value of y from the equation y = x - 3.
(c) solving the equation \(\displaystyle \, (2x\, +\, 3)\left(x\, +\, \sqrt{x\, +\, 3\,}\right)\left(x^2\, +\, x\right)\, =\, 0,\,\) and then deducing the value of y from the equation y = x + 3.
(d) solving the equation \(\displaystyle \, x^2\, -\, \sqrt{x\, +\, 3\,}\, -\, 25.11\, =\, 0,\,\) and then deducing the value of y from the equation y = x + 3.
(e) none of the above.

Other than the "to within" part, this is just algebra. What have you done with what you learned back in high school and in pre-calc?

Please be complete. Thank you!

 

[Ishuda]

PLEASE EXPLAIN:

Which 2 differences? Do you mean:
a = larger number + its square root
b = smaller number + its square
a - b = -66.33
?????

Where does the 25.11 come from? Do you mean 22.11?

"The difference between two numbers is 3" so let x be the smaller and we have
First difference = y-x = 3
Let a and b be given as above,
a = y + \(\displaystyle \sqrt{y}\) = x + 3 + \(\displaystyle \sqrt{x+3}\)
b = x + x²

Since the second difference wasn't specified, i.e. a-b or b-a, we have two cases to look at
>>>>>>>>>>>>>>>>HIGHLIGHTE FOR SOLUTION<<<<<<<<<<<<<<<<
Case (1)
Second difference = a-b = x + 3 + \(\displaystyle \sqrt{x+3}\) - (x + x²) = 3 + \(\displaystyle \sqrt{x+3}\) - x²
and
(y-x) (a-b) = 3 [3 + \(\displaystyle \sqrt{x+3}\) - x²] = -66.33
===> 3 + \(\displaystyle \sqrt{x+3}\) - x² = -22.11
===> \(\displaystyle \sqrt{x+3}\) - x² = -25.11
===> x² - \(\displaystyle \sqrt{x+3}\) - 25.11 = 0
Solve for x and compute y

OR
Case (2)
Second difference = b-a = -x - 3 - \(\displaystyle \sqrt{x+3}\) + (x + x²) = -3 - \(\displaystyle \sqrt{x+3}\) + x²
and
(y-x) (b-a) = 3 [-3 - \(\displaystyle \sqrt{x+3}\) + x²] = -66.33
===> -3 - \(\displaystyle \sqrt{x+3}\) + x² = -22.11
===> -\(\displaystyle \sqrt{x+3}\) + x² = -19.11
===> x² - \(\displaystyle \sqrt{x+3}\) + 19.11 = 0
Solve for x and compute y. How so some ever, that equation has no real solutions although there are (four?) complex ones.
>>>>>>>>>>>>>>>>HIGHLIGHTE FOR SOLUTION<<<<<<<<<<<<<<<<

 

[Ishuda]

This darn problem kept bugging me.....SO:

Let's use x = 9, y = 4
u = 1st difference = 5

2nd difference:
[x + SQRT(x)] - [x - u + (x - u)^2] = -8

v = product of differences = 5(-8) = -40

So u and v are the "givens", and we have:

u { [x + SQRT(x)] - [x - u + (x - u)^2] } = v; simplifying:

x^2 - 2ux - k = SQRT(x) , where k = u - u^2 - v/u

Squaring both sides and rearranging:
x^4 - (4u)x^3 + (4u^2 - 2k)x^2 + (4uk - 1)x + k^2 = 0

Substituting the givens:
x^4 - 20x^3 + 124x^2 - 241x + 144 = 0

Wolfram gives 3 solutions (not 4):
x = 9 : YAHOO!
x = ~1.2430
x = ~1.5728
http://www.wolframalpha.com/input/?i=x^4+-+20x^3+++124x^2+-+241x+++144+=+0

So I'm somewhat confused (what else is new!):
why is this thread named "numeric methods"?
how can a quartic have 3 solutions (not 4)?

Awaiting your words of wisdom, Ishuda (or anybody...)

Without going through this, I may later, the only explanation I see is that one is a double root.

EDIT:
First, I believe you have gone wrong above, see the red.

Next, went over to Wolframalpha and it seems to give a complex conjugate pair plus a pair at 8 plus a bit and 9. Also, when I look at the quartic you have above and put it in Excel, I find 4 roots: one between 1.2 and 1.3; one between 1.5 and 1.6, one between 8 and 8.5, and one at 9. Maybe the Excel table will copy over:

	f=	x^4 - 20x^3 + 124x^2 - 241x + 144
0.1	g=	x^3-11x^2+25x-16
x	f	g	g(x-9)-f
1.00000	8.00000	-1.00000	0.00000
1.10000	3.78410	-0.47900	0.00000
1.20000	0.87360	-0.11200	0.00000
1.30000	-0.82390	0.10700	0.00000
1.40000	-1.39840	0.18400	0.00000
1.50000	-0.93750	0.12500	0.00000
1.60000	0.47360	-0.06400	0.00000
1.70000	2.75210	-0.37700	0.00000
1.80000	5.81760	-0.80800	0.00000
1.90000	9.59210	-1.35100	0.00000
2.00000	14.00000	-2.00000	0.00000
2.10000	18.96810	-2.74900	0.00000
2.20000	24.42560	-3.59200	0.00000
2.30000	30.30410	-4.52300	0.00000
2.40000	36.53760	-5.53600	0.00000
2.50000	43.06250	-6.62500	0.00000
2.60000	49.81760	-7.78400	0.00000
2.70000	56.74410	-9.00700	0.00000
2.80000	63.78560	-10.28800	0.00000
2.90000	70.88810	-11.62100	0.00000
3.00000	78.00000	-13.00000	0.00000	0.5
3.50000	########	-20.37500	0.00000
4.00000	########	-28.00000	0.00000
4.50000	########	-35.12500	0.00000
5.00000	########	-41.00000	0.00000
5.50000	########	-44.87500	0.00000
6.00000	########	-46.00000	0.00000
6.50000	########	-43.62500	0.00000
7.00000	74.00000	-37.00000	0.00000
7.50000	38.06250	-25.37500	0.00000
8.00000	8.00000	-8.00000	0.00000
8.50000	-7.93750	15.87500	0.00000
9.00000	0.00000	47.00000	0.00000
9.50000	43.06250	86.12500	0.00000

Through for now

 

[Deleted member 4993]

The fourth '0' is at x = 8.18421

Mr. Newton told me that.....

 

[Deleted member 4993]

Ahhhh....but you didn't need this fellah Newton...

We had x = 9, 1.243 and 1.573

144 / (9 * 1.243 * 1.573) = 8.184

But then I would have to trust your calculations!!

Knowing how passionately you hate fractions, and those egregious mistakes of missing brackets - do you think that trust would be prudent??