Quadratic Equation
Solve the equation using the quadratic formulae.
3x -5x +7.
I have tried it and I came up b^2 - 4ac with square root of -59 and I stuck. Can we do a square root of a negative number?. The above question is right. Thanks.
Solve the equation using the quadratic formulae.
3x -5x +7.
I have tried it and I came up b^2 - 4ac with square root of -59 and I stuck. Can we do a square root of a negative number?. The above question is right. Thanks.
Rectify, the quadratic equation is 3x^2 - 5x = 7 . I appologise Thanks
You certainly are on the right track.
Yes, we can do a square root of a negative number. Your solution for x is an imaginary number and the answer is I*sqrt(59) where I^2 = -1.
Your equation needs to equal zero on one side before the quadratic equation will work. It has to do with the way the quadratic equation is derived, and I don't have time or tools to provide the details. Check out a video on deriving the quadratic equation if you are interested (there are dozens of good explanations on the internet). If you just need to find the answer, make the right side equal to zero by subtracting 7 from both sides.
Greetings to both of you. Thank you very much for the reply I apologise again for wrong typing of the question. In fact it is
3x^2 - 5x + 7 . This question involve complex root, but I really want yo know how to do it. I've reached tto the step of square root of -ve 59 for b^2 - 4ac and after that does it become square root of 59i where i is square root of -ve 1. Can I leave my final answer as such:
x = 5 + square root of 59i / 6 and/or
x = 5 - square root of 59i / 6 Briefly just to let you know that I'm not studying in any institution, I'm doing a course on line and I'm depending on myself and those who want to help me on line.Thanks.
When you get a chance, look up imaginary and complex numbers. There is a video in pre-calculus to help you understand the complex numbers.
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