Question 409319: Would you help me figure out this problem? I seem to mess up on signs. This is not coming out to be a perfect square, so I must have done somthing wrong.
Solving by quadratic formula -b sqrt (b^2-4ac)/(2a)
(3n-1)^2+2=18
(3n-1)^2+2-18-18
(3n-1)^2-18=0
3n^2+1+2-18=0
3n^2-(-15)=0 Did not see a b, so b is 1, right?
-1 sqrt ((1^2-4(3)(-15))/(2(3))
-1 sqrt sqrt (1-(-180)/(6)
-1 sqrt 181 I do not know what to do here, would you show me?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! The discriminant does not have to be a perfect square. Also, you expanded  incorrectly. We have
 Expand first.
 Gather up constant terms, move them to one side
For simplicity, I'll divide both sides by 3.
 Here, we have a = 3, b = -2, c = -5 (also, if there is no x term, b is 0, not 1). By the quadratic formula,
 = -1 or 5/3. Here, the discriminant turned out to be a perfect square, but it doesn't have to be the case. In fact, the discriminant doesn't even have to be nonnegative, since you can have nonreal roots as well.
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