91. If there is no real number solution to the quadratic equation, x^2+2x+C=0, what is a possible value of c?
1 answer:
For no real solution, you must have b^2 - 4ac of the quadratic formula be less than 0.
a = 1; b = 2; c = C
b^2 - 4ac < 0
2^2 - 4(1)(C) < 0
4 - 4C < 0
-4C < -4
C > 1
The only choice greater than 1 is 3.
Answer: the 4th choice, 3
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1 + -3 1/2 is = to -2.5 so v= -2.5
Check the solution in the attachment below.
2x+x-2=-2-x-1
3x-2=-3-x
4x=-1
x=-0.25
Answer:
<em>17 / 8</em>
Step-by-step explanation:
sec A = 
cos A = 
= 
⇒ <em>sec A = </em>
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