Use Descartes' Rule of Signs to explain why 2x^4 + 6x^2 + 8 = 0 h...

Question

Use Descartes' Rule of Signs to explain why 2x^4 + 6x^2 + 8 = 0 has no real roots.

Accepted Answer

Hi there. So this problem tells us to consider the following polynomial next to the fourth plus three X squared plus six and explain why it has no real roots. So the hit it gives us is to tell us to use the cards rule of science. The cars follow signs states that if we have any size changes in any of our terms here we will have a positive or a negative real route. Let's check to see if we have a positive or negative real root. So we have nine X to the fourth plus three X squared plus six. So if you remember for our book, we're gonna check to see if we have any sign changes. I'm not one to imagine first that there is a positive in front of the night or a plus. Now I'm going to check our sign changes. Well we have a plus here two here 2 here they're all plus no sign changes. If we have no sign changes, that means we have no positive rail routes at least no sign changes for positive. Those checks are negative. So it's a car slow signs, we'll plug in negative X everywhere where we have an X. So we did that were negative. We have nine negative X to the fourth plus three negative X squared plus six and want to see if we get a sign change this time. Well Native X the fourth is going to be positive X to the fourth. Native X squared it's positive squared and plus six we get the same as the positive answer but we have no sign changes again. No sign change. So there are no negative real rates as well. The offers are imaginary, which means we have no sign changes, which is we have answer A There are no sign changes between the terms when expecting for positive rail routes and negative fill roots. Okay. I hope will help you solve the problem. Thank you for watching. Goodbye.

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