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Geometry of Triangles

J

jarrhead

May 23, 2009
2,729
2
#1
I need help with my last question on my homework.


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If R = 2x2 - 7, S = 4x2 - 6x + 40, and T = 18x + 3, what is the measure of R?​


Anybody care to explain this?

I would think it'd be something like:

2x^2 - 7 + 4x^2 - 6x + 40 +18x + 3

6x^2 + 12x + 36 = 180
6x^2 + 12x = 144


What do I do from here?
 
#15
standupright said:
so R = 2 times the square root of 2 minus 7?
Click to expand...

if that's the case, then good luck.

inout.gif
 
#16
jarrhead said:
I know we have to find what x equals. Is there some form other than guess and check for:

6x^2 + 12x = 144
Click to expand...

I am not going to attempt to bewilder anyone , i am not smart with math and just barely passed math 111 with a 81%.

BUT

triangles are solved by Pythagereom theorum (i know i did not spell it right) which is a^2+b^=c^2 it seems that you are not putting 12x squared like 12x^2.
 
#18
jarrhead said:
6x^2 + 12x - 144 = 0

6x^2 - 144 = 12x
6x^2/12 - 12 = x

Im lost again.
Click to expand...

Factor 6x^2 + 12x - 144 = 0. You should end up with two first power binomials multiplied together (ax + b)(cx - d)=0 (one will have a plus and the other a minus because the constant in the original equation is negative), and then you can set each binomial pair equal to zero and solve them individually. Since the signs are different, one won't work in the original equations for the angles (it'll prododuce a negative angle value or something somewhere), so the other one will be what you go with.

If it doesn't factor (I think it does but I'm not sure, I've never been good at factoring trinomials), you can use the quadratic formula which is (-b +/- sqrt(b^2 - 4ac))/2a, and you'll get your 2 x values (if the equation doesn't factor they'll be radicals).
 
#20
Disraeli Gears said:
Factor 6x^2 + 12x - 144 = 0. You should end up with two first power binomials multiplied together (ax + b)(cx - d)=0 (one will have a plus and the other a minus because the constant in the original equation is negative), and then you can set each binomial pair equal to zero and solve them individually. Since the signs are different, one should come out to a negative and thus will be invalid for your purposes (can't have a negative angle measure).

If it doesn't factor (I think it does but I'm not sure, I've never been good at factoring trinomials), you can use the quadratic formula which is (-b +/- sqrt(b^2 - 4ac))/2a, and you'll get your 2 x values (if the equation doesn't factor they'll be radicals).
Click to expand...

exploding-head.jpg


We have to have 10 problems and an hour of time on study island. Four sections.

So, basically, I have 50 problems in the other three and 9 in this one. This is the last problem I need, and if I miss it, the my grade will drop to a 90 because I won't get the blue ribbon. ;)
 

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