anonymous
  • anonymous
I need to find the zeroes of f(x) = 5x2 + 2x + 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@AliceCullen @ramseysa @Elsa213 @DaBest21 @thomaster
BloomLocke367
  • BloomLocke367
zeroes are just your solutions. It's where x=0, or where the graph crosses the x-axis. Do you know how to solve quadratics?
anonymous
  • anonymous
No @BloomLocke367

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BloomLocke367
  • BloomLocke367
Here, take a look at this tutorial to see if you understand. If you still need help, please do tag me and I'll try my best to help you! http://openstudy.com/users/bloomlocke367#/updates/55acffcee4b071e6530c96ce
anonymous
  • anonymous
Wait, is -1/5 and 4/5
anonymous
  • anonymous
Are those the zeroes?
BloomLocke367
  • BloomLocke367
Actually, I'm pretty sure the zeroes are complex roots. One moment. I have to double check my work.
anonymous
  • anonymous
- 1/5 and - 4/5
BloomLocke367
  • BloomLocke367
No, those are not the roots. They are, in fact, complex. Do you know what a complex number is?
anonymous
  • anonymous
No
anonymous
  • anonymous
- 1/5 and - 2i/5
BloomLocke367
  • BloomLocke367
Basically, a complex number is a number that has an imaginary number in it. The often have a coefficient, i, and a constant. Here's an example of what it would look like: 3i+8. That's just an example, not the answer. Do you understand imaginary numbers? Where did you get those answers?
anonymous
  • anonymous
Mathway, I suck at math and despise it. I'm sorry, I'm a history and English guy.
anonymous
  • anonymous
Actually, I got them from Wolfram/Alpha
BloomLocke367
  • BloomLocke367
Really? huh. Try using the quadratic formula, which is \(\LARGE x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
BloomLocke367
  • BloomLocke367
In this case, a is 5, b is 2, and c is 1.
anonymous
  • anonymous
(-2±√(2^2-4*5*1))/(2*5)
anonymous
  • anonymous
Is it 1/10(-2 ± 4i)?
anonymous
  • anonymous
@badmood
radar
  • radar
Actually you were right. You stated you had to find the 0"s of f(x) = 5x^2 + 2x + 1 This is the value where f(x) = 0. Obviously f(x) depends on x and x can be an infinite number of values both positive and negative and complex. When x = 0, in this given function f(x) = 1, when x = -1, f(x) = 4 etc. for f(x) = 0 you would solve for x, and the quadratic equation would be a reasonable method, however when you graph this function you will find that the f(x) nevers gets to 0. This does not mean that the function does not exists it simply means there is nor real "0" for f(x) for any real value of x.

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