SOLUTION: 1. Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer. a) x2 + 3x - 15 = 0 b) x2 + x + 4 = 0 c)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 1. Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer. a) x2 + 3x - 15 = 0 b) x2 + x + 4 = 0 c)       Log On


   

Question 199098: 1. Determine the number of solutions and classify the type of solutions for each of the following equations. Justify your answer.
a) x2 + 3x - 15 = 0
b) x2 + x + 4 = 0
c) x2 – 4x - 7 = 0
d) x2 – 8x + 16 = 0
e) 2x2 - 3x + 7 = 0
f) x2 – 4x - 77 = 0
g) 3x2 - 7x + 6 = 0
h) 4x2 + 16x + 16 = 0
any help would be appriciated.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first three to get you started (they all follow the same logic)

a)


From x%5E2%2B3x-15 we can see that a=1, b=3, and c=-15


D=b%5E2-4ac Start with the discriminant formula.


D=%283%29%5E2-4%281%29%28-15%29 Plug in a=1, b=3, and c=-15


D=9-4%281%29%28-15%29 Square 3 to get 9


D=9--60 Multiply 4%281%29%28-15%29 to get %284%29%28-15%29=-60


D=9%2B60 Rewrite D=9--60 as D=9%2B60


D=69 Add 9 to 60 to get 69


Since the discriminant is greater than zero, this means that there are two real solutions.



b)



From x%5E2%2Bx%2B4 we can see that a=1, b=1, and c=4


D=b%5E2-4ac Start with the discriminant formula.


D=%281%29%5E2-4%281%29%284%29 Plug in a=1, b=1, and c=4


D=1-4%281%29%284%29 Square 1 to get 1


D=1-16 Multiply 4%281%29%284%29 to get %284%29%284%29=16


D=-15 Subtract 16 from 1 to get -15


Since the discriminant is less than zero, this means that there are two complex solutions.


In other words, there are no real solutions.




c)




From x%5E2-4x-7 we can see that a=1, b=-4, and c=-7


D=b%5E2-4ac Start with the discriminant formula.


D=%28-4%29%5E2-4%281%29%28-7%29 Plug in a=1, b=-4, and c=-7


D=16-4%281%29%28-7%29 Square -4 to get 16


D=16--28 Multiply 4%281%29%28-7%29 to get %284%29%28-7%29=-28


D=16%2B28 Rewrite D=16--28 as D=16%2B28


D=44 Add 16 to 28 to get 44


Since the discriminant is greater than zero, this means that there are two real solutions.