3x^2-7x-3=0

- anonymous

3x^2-7x-3=0

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- katieb

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- amistre64

3times3 is 9 which doesnt factor into 7.... so quadratic it :)

- anonymous

after using the quadratic formula, how do you complete the problem?

- amistre64

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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## More answers

- anonymous

my textbook says to find the nearest degree.. x = sin of theta

- amistre64

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?

- anonymous

not really :/

- amistre64

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?

- anonymous

ok, let me see.

- anonymous

where do you get 49?

- amistre64

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

- anonymous

oh. it's b^2? no wonder i never got any solution out of that. thank you very much!
+5

- amistre64

we got numbers for our x now; (7/6 (+-) sqrt(85)/6) if I did it correctly right?

- anonymous

umm.. i'm still solving it. i got the b^2 correct

- anonymous

my calculator says -49 instead of 49. so weirdï½ž

- amistre64

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.

- anonymous

i think it was supposed to be 7 not -7

- amistre64

all it changes in the end is the sign of (7/6) to (-7/6) which may or may not be an issue

- anonymous

got it.
\[x=7\pm \sqrt{49+36} \div 6\]

- amistre64

if your equation you gave was correct, then it was supposed to be (-7)^2 = 49. But (7)^2 = 49 just the same

- amistre64

good :) you ready to learn about sin inverse? its symbol is (sin^-1)

- anonymous

i know sin^-1 but i'm stuck in where should i go next.

- amistre64

lets walk it thru. It asks us to find sin(t) = x right?

- anonymous

sin \[\theta\]

- amistre64

theta is simply a name, I will make it simpler and just name it "t" do you think I can do that?

- anonymous

yea, sure. :)

- amistre64

lol...good, mathmaticians like to rename things all the time, makes them feel important :)

- anonymous

XD

- amistre64

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

- anonymous

wait. i'm confused here. do i continue with the quad. formula?

- amistre64

the quad formula simply gave us the values for x....its job is done and it can go take a well deserved break...

- anonymous

so i continue with this:\[3\sin^2\theta-7\sin \theta-3=0\]

- amistre64

aha!!! you were holding back vital information...hmmm

- anonymous

opps.. am i in trouble?

- amistre64

nah....just makes the problem more understandable now :)

- anonymous

XD. so how do i continue?

- amistre64

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..

- anonymous

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.

- amistre64

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.

- amistre64

[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.

- amistre64

[7-sqrt(85)]/6 = -0.3699 which IS between -1 and 1 so it is a good value of x.

- amistre64

is there anyway to verify the answer?

- anonymous

i got the same answer as you. -0.3699240762

- amistre64

either were both geniuses or were both idiots...take your pick :)

- anonymous

GENIUS! :D

- amistre64

yay!!!

- amistre64

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.

- amistre64

does any of that make sense to you?

- anonymous

somewhat, let me think over~

- anonymous

i think i got it, thanks for helping me though(:

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