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Mathematics 10 Online
OpenStudy (anonymous):

Of the following statements, how many are true? -A quartic polynomial can have exactly one imaginary root. -A cubic polynomial must have a real root. -A quadratic polynomial can have no real roots -A polynomial can have no real roots

OpenStudy (anonymous):

@FibonacciChick666

OpenStudy (fibonaccichick666):

So, what do you know about imaginary roots?

OpenStudy (anonymous):

they come in pairs

OpenStudy (fibonaccichick666):

exactly, so can 1 be true?

OpenStudy (anonymous):

I don't think that first one is true. I know the second and third one is true. but I'm not sure if the fourth one is true?

OpenStudy (fibonaccichick666):

correct, the first is false

OpenStudy (fibonaccichick666):

if a quadratic polynomial can have no real roots, why can't a generic polynomial?

OpenStudy (fibonaccichick666):

is a quadratic, not a polynomial?

OpenStudy (anonymous):

yes but would it look like this|dw:1456680747230:dw|

OpenStudy (fibonaccichick666):

yea, something like that

OpenStudy (anonymous):

so it doesn't have to have an real roots?

OpenStudy (anonymous):

so the 2, 3, and 4 statement are correct.

OpenStudy (fibonaccichick666):

yep

OpenStudy (fibonaccichick666):

take \((x^2+1)=0\) There are no real roots

OpenStudy (anonymous):

okay thank you very much. I get it now!

OpenStudy (fibonaccichick666):

but by the fundamental theorem of algebra, we know that there must be as many roots as the degree of the polynomial. So in that case, there are 2 complex(imaginary) roots

OpenStudy (fibonaccichick666):

np!

OpenStudy (anonymous):

If x^8 has only 2 real roots and x^9 has three real roots how does this work?

OpenStudy (fibonaccichick666):

well, the rest are either multiples or imaginary pairs

OpenStudy (fibonaccichick666):

note, I didn't say distinct roots, just that there are roots

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