3x^2-7x-3=0

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3times3 is 9 which doesnt factor into 7.... so quadratic it :)
after using the quadratic formula, how do you complete the problem?
that is the completed problem. you could try doing the "complete the square" method; but all that method does is prove that the quad formula works.

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my textbook says to find the nearest degree.. x = sin of theta
then find what x equals and take the sine inverse of it. the inverse trig funtions take numbers for inputs and spit back angles. Does that make sense?
not really :/
lets walk you thru it then :) first get the values for x with the quad formula. 7/6 (+-) sqrt(49-(4)(-3)(3))/ 6 7/6 (+-) sqrt(49+36)/6 7/6 (+-) sqrt(85)/6 you with me so far?
ok, let me see.
where do you get 49?
the quad form reads: -middle (+-) sqrt(middle^2 -4(first)(last)) all over 2(first). or more formally: -b (+-) sqrt(b^2-4(a)(c)) --------------------- 2(a) right? -7 times -7 = 49
oh. it's b^2? no wonder i never got any solution out of that. thank you very much! +5
we got numbers for our x now; (7/6 (+-) sqrt(85)/6) if I did it correctly right?
umm.. i'm still solving it. i got the b^2 correct
my calculator says -49 instead of 49. so weird´Ż×
lol.... maybe user error? To find out what sin(t) is when x equals some number we write it like this: sin(t) = x To find out what "t" is we need to get rid of the (sin) part. And we do that by using what is called the sine inverse function.
i think it was supposed to be 7 not -7
all it changes in the end is the sign of (7/6) to (-7/6) which may or may not be an issue
got it. \[x=7\pm \sqrt{49+36} \div 6\]
if your equation you gave was correct, then it was supposed to be (-7)^2 = 49. But (7)^2 = 49 just the same
good :) you ready to learn about sin inverse? its symbol is (sin^-1)
i know sin^-1 but i'm stuck in where should i go next.
lets walk it thru. It asks us to find sin(t) = x right?
sin \[\theta\]
theta is simply a name, I will make it simpler and just name it "t" do you think I can do that?
yea, sure. :)
lol...good, mathmaticians like to rename things all the time, makes them feel important :)
XD
we have: sin(t) = x we need to get rid of the "sin" part. So lets apply sin^-1 to each side to get: t = sin^-1(x) Now plug the numbers into your scientific calculator to get an angle for t, formally known as theta
wait. i'm confused here. do i continue with the quad. formula?
the quad formula simply gave us the values for x....its job is done and it can go take a well deserved break...
so i continue with this:\[3\sin^2\theta-7\sin \theta-3=0\]
aha!!! you were holding back vital information...hmmm
opps.. am i in trouble?
nah....just makes the problem more understandable now :)
XD. so how do i continue?
just the same as before :) sin(t) = x and solve for t But I gotta tell wonder, do you have the problem written correctly becuase I get some wierd stuff when I plug it into the calculator..
the textbook says to use the quadratic formula to find,to the nearest degree, all values of theta in the interval 0 degrees [less than or equal to] theta [less than] 360 degrees that satisfy each equation.
338.29 degrees is what I get, which is in Quad4 but it is also hiding in Quad3 under the disguise of 201.71 degrees.
[7+sqrt(85)] /6 is greater then 1 so it spits out an error. x = 2.7 and that aint between -1 and 1 so it is a false reading.
[7-sqrt(85)]/6 = -0.3699 which IS between -1 and 1 so it is a good value of x.
is there anyway to verify the answer?
i got the same answer as you. -0.3699240762
either were both geniuses or were both idiots...take your pick :)
GENIUS! :D
yay!!!
the thing about sin^-1 is that it only gives an angle that is on the right hand side of a circle. So you have to remember to look over to the other sid eto see where it matches up at.
does any of that make sense to you?
somewhat, let me think over~
i think i got it, thanks for helping me though(:

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