Help with equations

[branko917]

I need help for final solution for 3 different equations, i solved some of the steps but i get stuck and cannot find final solution.
Please help if you can, i need detailed explanation for each step. This is the third time i post this question and i hope that i did it right regarding to the moderator remark.
This is not a school homework, i am 40+ mechanical engineer attempting to remind of engineering mechanics-statics. Years past by and some things got forgotten.
Problems is in attachment.

Thanks.

 

[Steven G]

For eq 1 you put in the the wrong value for c into the quadratic formula.

 

[Otis]

Hi branko. For eq2, we'd need to square both sides (at least twice) to eliminate radicals. I'm thinking the resulting degree would be too high to solve algebraically with paper and pencil. Do you remember any numerical methods? (Sometimes, we need to use software.)

?

 

[Otis]

… This is the third time i post this question …

I looked at your threads, branko. You've posted three equations that are similar, but they're not the same.

For eq3, I don't understand your first step. Why did you change 3000z to 30z?

Here are my first four steps, for that equation:

Add 981 to each side
Divide each side by 3000
Square each side
Multiply each side by (0.25 + x^2)

I'll leave the last two steps for you to try and finish.

?

 

[branko917]

I looked at your threads, branko. You've posted three equations that are similar, but they're not the same.

For eq3, I don't understand your first step. Why did you change 3000z to 30z?

Here are my first four steps, for that equation:

Add 981 to each side
Divide each side by 3000
Square each side
Multiply each side by (0.25 + x^2)

I'll leave the last two steps for you to try and finish.

?

I made an error in first step in eq3. I reviewed your first four steps and found correct solution of the equation, z=0,173.
Thank you very much.

 

[branko917]

For eq 1 you put in the the wrong value for c into the quadratic formula.

Yes, i can see now, if i am right, i put positive value of 240590,25 instead of negative, but i still cannot obtain correct result which is F=519,79. My results is F1=495,5 and F2=-5384,5, quadratic equation, two results.

 

[branko917]

Hi branko. For eq2, we'd need to square both sides (at least twice) to eliminate radicals. I'm thinking the resulting degree would be too high to solve algebraically with paper and pencil. Do you remember any numerical methods? (Sometimes, we need to use software.)

?

I tried everything but i cannot get right value. Correct value is d=1,64, i know that because this is from one problem in statics, you must correctly form equilibrium equation which i did it right but i got stuck with calculation of final solution. This equation could be written in another form (picture in attachment). I tried to solve in this form but i got stuck after few steps.

 

[Steven G]

Yes, i can see now, if i am right, i put positive value of 240590,25 instead of negative, but i still cannot obtain correct result which is F=519,79. My results is F1=495,5 and F2=-5384,5, quadratic equation, two results.

Please try not to suggest that 240590,25 is a negative value as it is not.
Can we see your updated results so we can find your error?

 

[Otis]

I tried everything but i cannot get right value …

Hi branko. To confirm that we can't solve eq2 algebraically, I did the following to each side (using your first version):

Subtract 80
Divide by -3
Multiply by sqrt(d^2 + 2.25)
Square
Expand
Subtract 2500d^4 and 11250d^2
Square
Expand

At that point, I had an 8th-degree polynomial equation:

56250000*d^8 - 32000000*d^6 - 211448889*d^4 - 303520000*d^2 + 23040000 = 0

We don't have formulas for finding roots of polynomials above the fourth degree. To approximate by hand, we could use a root-finding algorithm (eg: Newton's Method). I used software, to confirm your solution. The 8th-degree polynomial above has four Real roots. Checking each, 1.6403 (rounded) is the only one that works. The other three solutions are extraneous. (Squaring an equation can lead to false solutions because signs are lost).

?

 

[Otis]

… i put positive value of 240590,25 instead of negative …

Yes, I see that sign error, but Jomo had missed an earlier issue: round-off error. When you combined f² terms, you got 0,086f². That coefficient is off because you'd rounded prior results too much.

We need to keep more decimal places. Sometimes, you've rounded numbers to two places. Sometimes, you've rounded numbers to one decimal place. When we round, we're throwing away (or adding) little bits of numbers. If we lose or add too much, our final result will be incorrect. You'd combined the f² terms like this:

f² - (0,734f² + 0,18f²) = 0,086f²

When we carry four decimal places, we get this instead:

f² - (0,7344f² + 0,1840f²) = 0,0816f²

Using four decimal places throughout, we get the solution 519,38 (rounding to two places only as the very last step).

If we use "all" decimal places (i.e., don't round calculator results; use all digits displayed), then we get the solution 519,42 (rounding to two places only at the very end).

You're expecting the solution 519,79 but where did your original numbers come from? Did you round too much, when calculating the beginning coefficients 0,857 and 490,5 and 0,429? If those coefficients were reported with more decimal places, then maybe we could get 519,79.

?

 

[Otis]

Please try not to suggest that 240590,25 is a negative value …

? I think branko's statement suggests that he understands the sign error.

For something completely different -- when we use commas as decimal points, do you know whether we use periods to separate decimal numbers in a list? For example, can we write "0,857 and 490,5 and 0,429" as:

0,857. 490,5. and 0,429

?

 

[Steven G]

? I think branko's statement suggests that he understands the sign error.

For something completely different -- when we use commas as decimal points, do you know whether we use periods to separate decimal numbers in a list? For example, can we write "0,857 and 490,5 and 0,429" as:

0,857. 490,5. and 0,429

?

No, I do not know that. What about a different list, like a grocery list?

 

[branko917]

Yes, I see that sign error, but Jomo had missed an earlier issue: round-off error. When you combined f² terms, you got 0,086f². That coefficient is off because you'd rounded prior results too much.

We need to keep more decimal places. Sometimes, you've rounded numbers to two places. Sometimes, you've rounded numbers to one decimal place. When we round, we're throwing away (or adding) little bits of numbers. If we lose or add too much, our final result will be incorrect. You'd combined the f² terms like this:

f² - (0,734f² + 0,18f²) = 0,086f²

When we carry four decimal places, we get this instead:

f² - (0,7344f² + 0,1840f²) = 0,0816f²

Using four decimal places throughout, we get the solution 519,38 (rounding to two places only as the very last step).

If we use "all" decimal places (i.e., don't round calculator results; use all digits displayed), then we get the solution 519,42 (rounding to two places only at the very end).

You're expecting the solution 519,79 but where did your original numbers come from? Did you round too much, when calculating the beginning coefficients 0,857 and 490,5 and 0,429? If those coefficients were reported with more decimal places, then maybe we could get 519,79.

?

When i rounded numbers to five places i got correct result F=519,79 N, solution with original numbers in attachment.
Rounding to more numbers gave me expected correct result.

Thanks for your help again.

 

[branko917]

Hi branko. To confirm that we can't solve eq2 algebraically, I did the following to each side (using your first version):

Subtract 80
Divide by -3
Multiply by sqrt(d^2 + 2.25)
Square
Expand
Subtract 2500d^4 and 11250d^2
Square
Expand

At that point, I had an 8th-degree polynomial equation:

56250000*d^8 - 32000000*d^6 - 211448889*d^4 - 303520000*d^2 + 23040000 = 0

We don't have formulas for finding roots of polynomials above the fourth degree. To approximate by hand, we could use a root-finding algorithm (eg: Newton's Method). I used software, to confirm your solution. The 8th-degree polynomial above has four Real roots. Checking each, 1.6403 (rounded) is the only one that works. The other three solutions are extraneous. (Squaring an equation can lead to false solutions because signs are lost).

?

I solve good part of eq2, it all comes down to a fourth-degree reciprocal equation that must be come down to quadratic equation and final solution, i am not capable to solve any further. I will keep trying.
Partial solving method in attachment.
Do you think that this is the right direction to the final solution.

 

[JeffM]

There is a very ugly formula for finding the roots of polynomials of the fourth degree. Or you can use the Newton-Raphson method. Or you can use good graphing software. If it is virtually certain that you are not going to get a rational answer so approximation is just fine.