1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Minimum from a completion of a square

  1. Oct 31, 2005 #1
    so the question is:

    Complete the square of x^2 + 4x - 1 and hence find the position and value of its minimum.

    AFter completing the square i have : (x+2)^2 - 5 ,

    How do I find the minimum value from that?

    Thanks in advance guys!
     
  2. jcsd
  3. Oct 31, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    What is the smallest possible value for [itex](x+2)^2[/itex]?
     
  4. Oct 31, 2005 #3
    umm so would x= -2 and y = -5

    ??
     
  5. Oct 31, 2005 #4

    TD

    User Avatar
    Homework Helper

    Indeed :smile:
     
  6. Oct 31, 2005 #5
    okaay another questoin....
    Factorise and hence solve the equation:

    2x^3 + 3x^2 - 8x - 12 = 0

    I can't seem to remember how to factorize this because there's a cubic expression! how do i start?
     
  7. Oct 31, 2005 #6

    TD

    User Avatar
    Homework Helper

    Divisors of the constant term (i.e. -12 here) are possible zeroes of your equation. Try to find such a zero and use the fact that if x = a is a zero, you can factor out (x-a).
     
  8. Oct 31, 2005 #7
    okay thanks, i'm going to try that now, by the way is that what you call the remainder theorem?
     
  9. Oct 31, 2005 #8
    Okay, so my answer was:

    (x-2)(x+2)(2x+3)

    is that right?
     
  10. Oct 31, 2005 #9

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    You can expand that product and see if it produces your original expression.
     
  11. Oct 31, 2005 #10
    So there's this question to factorize:

    x^4 - 3x^2 - 10 = 0

    How do I start this one because I can't find a factor to go into it.
     
  12. Oct 31, 2005 #11

    TD

    User Avatar
    Homework Helper

    Yes, which is easy to see if you follow Tide's advice.

    Would it be easier for you to see if you let [itex]x^2 = t[/itex] so it becomes quadratic?

    That would give [tex]t^2-3t-10=0[/tex] to factor and then substitute t by x² again.
     
  13. Oct 31, 2005 #12
    so the ans is [tex]\sqrt{5}[/tex] ?
     
    Last edited: Oct 31, 2005
  14. Oct 31, 2005 #13

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    That's one of them! :)
     
  15. Oct 31, 2005 #14
    Ohh is it +/- Square root 5?! lol
     
  16. Oct 31, 2005 #15

    TD

    User Avatar
    Homework Helper

    Indeed, unless you're working complex since then a quartic equation has 4 solutions. The ones you found are the only real ones though.
     
  17. Oct 31, 2005 #16
    help with this last one

    Express the function [tex]\sqrt{3}\sin{2t} - 3\cos{2t}[/tex] in the form A[tex]\sin{(2t+\alpha)[/tex]


    i have no idea what to do
     
    Last edited: Oct 31, 2005
  18. Oct 31, 2005 #17

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Now you're just guessing!

    [tex]t^2 - 3t - 10 = (t-5)(t+2)[/tex]
     
  19. Oct 31, 2005 #18

    TD

    User Avatar
    Homework Helper

    Yes, but t was x².
     
  20. Oct 31, 2005 #19
    well if you sub the [tex]x^2[/tex] back in to the equation i get [tex]\pm\sqrt{5}[/tex]. is that not correct? and what of the other question? how should i start what am i suppose to do?
     
  21. Oct 31, 2005 #20

    TD

    User Avatar
    Homework Helper

    I would 'expand' sin(2t+a) with the formula for addition of angles, being:

    [tex]\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b[/tex]
     
  22. Oct 31, 2005 #21
    ya i tried to do that but i did not equal the left side.... so....? am i doin something wrong?
     
  23. Oct 31, 2005 #22

    TD

    User Avatar
    Homework Helper

    It won't be equal to it immediately but some parts will be the same. Identify the other factors with the ones which were there in the given expression.
     
  24. Oct 31, 2005 #23
    ok im completely lost now heres what i got:

    [tex]\sin{2t}\cos{\alpha} + \cos{2t}\sin{\alpha}[/tex]

    theres nothing more that i can do to make it look like the right side.
     
  25. Oct 31, 2005 #24

    TD

    User Avatar
    Homework Helper

    Don't forget the factor a that was in front of that sine at the RHS.
    Then compare the two terms with those on the LHS. What is already equal and which parts do you have to 'make' equal?
     
  26. Oct 31, 2005 #25
    ??how do i factor A ?i dont see anything that is equal on the RHS...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook