Use the Quadratic formula to find the roots, which will be the answer.
for ax^2 + bx + c = 0
x =( -b +|- sqrt(b^2 - 4ac))/2a
a = 1 b = 18 c = 36
x = (-18 +|- sqrt(18^2 - 4(1)(36)))/2(1) x = (-18 +|- sqrt(180))/2 x = -9 +|- 3/2*sqrt(20) x = -9 + -1.5*sqrt(20) and x = -9 + 1.5*sqrt(20) x is about -15.7 and -2.3
Check x = -9 + 1.5*sqrt(20) by substituting into original equation (-9 + 1.5*sqrt(20)+ 6)(-9 + 1.5*sqrt(20) + 12)= (-3 + 1.5*sqrt(20))(3+ 1.5*sqrt(20)) = -9 + 4.5*sqrt(20) + -4.5*sqrt(20) + (1.5)(1.5)(sqrt(20))(sqrt(20))= -9 + 45 = 36
Check x = -9 + -1.5*sqrt(20) by substituting into original equation (-9 + -1.5*sqrt(20) + 6)(-9 + -1.5*sqrt(20) + 12)= (-3 + -1.5*sqrt(20))(3 + -1.5*sqrt(20))= -9 + 4.5*sqrt(20) + -4.5*sqrt(20) + (-1.5)(-1.5)(sqrt(20))(sqrt(20))= -9 + 45 = 36
Thanks for all the fun math problems. I don't get to do this kind of crap any more being a stay at home mom.
I just noticed that I could have written 3/2sqrt(20) as 3sqrt(5). That would have been a lot simplier, and easy to read. It doesn't really matter anyhow, since they are the same number. I doubt anyone else cares either. :)
Augmented matrices are especially useful when solving simultaneous equations. There are tons of applications for simultaneous equations in engineering and computer science (3D Graphics, Robotics, etc.)
I took electrical engineering courses too. I was a computer engineering major, which is kind of a cross between electrical engineering and computer science.
That's what CE ended up being at my school... I kind of wish I'd gone the route of CS/CE, now. I got into database stuff/programming after I left school, and I really enjoy that.
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