Asked by Mitch n' Joey
1. How many real solutions does the function shown on the graph have?
The graph is following the function: y=(x+2)^2
A. No real solutions
B. One real solution** My answer
C. Two real solutions
D. Cannot be determined
2. What is the solution to 3x^2+3x+5=0?
I have the answers down to either -3(+or-)i squareroot51 / 6 or 3(+or-) i squareroot51 / 6
(I have to use the quadratic function)
Thank you so much!
The graph is following the function: y=(x+2)^2
A. No real solutions
B. One real solution** My answer
C. Two real solutions
D. Cannot be determined
2. What is the solution to 3x^2+3x+5=0?
I have the answers down to either -3(+or-)i squareroot51 / 6 or 3(+or-) i squareroot51 / 6
(I have to use the quadratic function)
Thank you so much!
Answers
Answered by
Reiny
I agree with #1
The way you typed the answer to #2 would indicate 4 answers, but a quadratic has only 2 solutions.
Be careful how you use the quadratic formula:
for ax^2 + bx + c = 0
x = (-b ± √(b^2 - 4ac) )/(2a)
in your case:
x = (-3 ± √-51)/6
= (-3 ± √51 i)/6 , and that's all
the ± only applied to the square root part.
The way you typed the answer to #2 would indicate 4 answers, but a quadratic has only 2 solutions.
Be careful how you use the quadratic formula:
for ax^2 + bx + c = 0
x = (-b ± √(b^2 - 4ac) )/(2a)
in your case:
x = (-3 ± √-51)/6
= (-3 ± √51 i)/6 , and that's all
the ± only applied to the square root part.
Answered by
Mitch n' Joey
Yes, sorry about that! I didn't know how to type in the necessary symbols. I was only confused concerning the 3 in the final answer. I wasn't sure if it was meant to be -3 or 3. From my understanding, I was right in thinking it was -3, right? :)
Answered by
Reiny
yes, look at my final answer
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