Results 1 to 2 of 2

Math Help - Quadratics ad other stumpers

  1. #1
    Newbie
    Joined
    Aug 2005
    Posts
    16

    Quadratics ad other stumpers

    Can someone explain quadratic formula to me here are the problems I am having trouble with:

    V^2+8v+6=o

    -x^2-5x+1=0

    2t^2-6t+1=0


    I don't know where to begin, also when I'm trying to complete a square such as:

    2m^2-m-15=0
    where do I start to complete equations is
    2x^2+5-1=5

    if I am trying to solve this type of equation:

    2x^2-1=x squared and it is telling me to look for extraneous solution what is it trying to tell me to look for and where do Ibegin with this type

    If I am looking for a perfect square :
    3x^2+15+9 do I not need to filter out the GCF forst and make the equation
    3(x^2+5+3) ----- and if I do that there is no perfect square please inform

    Trying to solve with formulas where would you start:

    Example ---- v=4/3(pi)r^2h and I'm trying to solve for h where again do I begin

    Solving proportions:

    -5/9=3/x or 9/3=3/2

    am I correct in saying that the answers are -----x= -27/5 and x= 6

    If i have 2m/3 = 3m/2
    Is my answer:
    4m=9m
    9m-4m=0
    5m=0
    m=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Aug 2005
    Posts
    53
    Hi, first if you have many question you should post different posts for every type of questions.

    1) V^2+8v+6=0, -x^2-5x+1=0, 2t^2-6t+1=0 with the quadratic formula.
    ok, so a quadratic equation is an equality with (in your case) one variable having powers of at most 2. So we can write it as a*x^2 + b*x+c=0 where a,b,c are known numbers and b and/or c can equal 0 (if a=0 then the x^2 will disappear and it won't be quadratic anymore). So you learn that to find the values of x satisfaying the equation ax^2+bx+x=0 you use [-b +- sqrt(b^2 - 4ac)]/(2a) where the +- means that use use this formula twice : once with + and you find the 1st possible value of x (x1) and then you put - and find the 2nd possible value of x (x2).

    So for example : v^2 + 8v + 6 =0 then : a=1,b=8 and c=6 so x1=[-b + sqrt(b^2 - 4ac)]/(2a) = [-8+sqrt(40)]/2 = -0.8377 we can call it v1 since your variable is v so v1=-0.8377. By the way, if you replace v1 in v^2+8v+6=v1^2+8*v1+6 you'll find 0 (That is the point of finding the values v1 and v2). and v2 = [-8-sqrt(40)]/2 = 0.8377 (same but we replace the "plus" by a "minus")=-7.1622 and again if you replace v2 in your equation you'll find 0.

    You do the same with the other two (caution : the a in the second is -1)

    For your personal information : there are usually two values for the variable that satisfy a quadratic equation that is because the graph (parabola) of the quadratic equation usually cuts the x axis 2 times and those points of intersection with the x axis are the values you look for.

    2) complete square : 2m^2-m-15=0. See 4).

    3) "2x^2-1=x squared " ??? is the x at the right squared ? but x squared = x^2 so you have x^2 in both sides. I don't know what is extraneous solutions but if the equation is 2x^2 - 1 = x^2 then the solutions are

    2x^2 - x^2 = 1 so x^2 = 1 so x=1 or x=-1 (both x1 and x2 squared give 1)
    _________________________________________________
    If the equation was 2x^2-1=x so you put x on the other side by changing its sign : so it becomes -x and we get 2x^2-x-1=0 and then you solve as in 1) and get x1=1 et x2=-0.5.

    4) for the perfet square thing : a perfect square is (a*variable + b)^2. Where a and b are not roots. example I can put 2x^2 as (sqrt(2)x)^2 but it is not beautiful. In the contrary, 4x^2 as (2x)^2 is better so we look for squares like those to do our perfect square.

    your example : 3x^2+15x+9. Here 3 is not a square. So we better remove it right away even if it is not a gcd of the others. so we have 3(x^2+5x+3). So now, if we had a square (ax+b)^2 it would equal (ax+b)(ax+b)=a^2 x^2 + 2axb + b^2 so in x^2+5x+3 a^2=1 and 2axb = 5x so b=5/(2a)=5/2. So we need the b^2 to complete our square. since b=5/2, b^2=25/4 so x^2+5x+25/4 is a perfect square and = (ax+b)^2 = (x+5/2)^2 but we have a 3 instead of the 25/4 so we remove what we want from the 3 and leave the waste. 3=25/4+? so ?=3-25/4=-13/4. So 3(x^2+5x+3) = 3(x^2+5x+25/4 - 13/4) = 3([x^2+5x+25/4]-13/4) = 3((x+5/2)^2 - 13/4) = 3(x+5/2)^2 - 39/4 (I removed the big parentheses so I multiplied both (x+5/2)^2 and -13/4 by 3 ).

    5) v=4/3(pi)r^2h. This is a formula in more than one variable. example : solving y=2x in x is x=y/2 and solving x=y/2 for y is y=2x. When you say I want to solve an equation in a variable (let's say f) it means you want to put f=... (you isolate f et get f=...) as we did for x and y. To do that you must begin by removing what is (in the sense of priority of operations) the one that has the least priority first and so on until you end up with just that variable alone on one side. In the first example, it was easy since we just had one thing only with x. But if we had to solve for x the following :y=2x+1 then you know addition (+1) has less priority then multiplication (2*) so we begin by removing the addition with its inverse operation (subtraction) so we subtract one from both sides and get : y-1 = 2x + 1 - 1 = 2x so y-1=2x. Now, since 2 multiplies x and the inverse of mutiplication is division, we divide both sides by 2 and get (y-1)/2=x and done ! x= (y-1)/2. In your example, all 4/3 and pi and r^2 mutilply h so they all have the same priority regarding h so to solve for h is just to do the contrary of the multiplication for all those factors and divide them in both sides ( we can do them all together since they have the same priority) and get v /(4/3 pi r^2)=h which is the same as h=3v/(4 pi r^2).

    6) Solving proportions : the first one : -5/9=3/x : so -5x = 27 so x= -27/5 good ! and the second one you give is : 9/3=3/2 (I think you missed an x somewhere since 9/3 is not equal to 3/2) Just tell me where is your x.

    7) If you have 2m/3 = 3m/2. yes m=0 no problem.

    (I am in a hurry now but if I have time later, I will give some other advices so you should maybe check it again some other time)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help me on these all night stumpers
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: August 28th 2009, 02:34 PM
  2. Quadratics
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 8th 2008, 01:39 AM
  3. Quadratics
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 15th 2008, 04:04 PM
  4. Help with Quadratics
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 19th 2008, 08:09 PM
  5. Quadratics, Help!
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 17th 2008, 04:37 PM

Search Tags


/mathhelpforum @mathhelpforum