Under what circumstances would a quadratic inequality have an empty solution set?

Question

Thanks for the help.

Answer 1

If the solution to the equality were imaginary. For example, if you were given x^2+5x+4<0. Since there are no values of x that would produce a result that is less than zero the solution would be the empty set.

Answer 2

a million: the variety of a polynomial is related to its degree because of the fact the polynomial is bounded on one component, i.e. won't bypass to infinity, if the polynomial is "even", which ability the suitable means is a good quantity, like 2,4,6 and so on. no rely if this is atypical, it is going to bypass from adverse infinity to valuable infinity. 2: seek for the place the graph crosses the x-axis, and those are your zeros. 3: sure, because of the fact 0 situations something is 0. 4: No because of the fact an nth degree polynomial can basically have as much as n aspects that incorporate the variable, and there is largely one 0 in line with component. 5: No, using intermediate fee theorem. All numbers between an larger certain (infinity) and a decrease certain (adverse infinity) will take place quicker or in a protracted time the graph, and because 0 is between infinity and adverse infinity, a third degree polynomial would desire to have a root. 6: Horizontal asymptotes and oblique asymptotes the two clarify a graph because it is going to infinity, and it does no longer be a function if it is going to 2 distinctive values. 7: it might desire to have a closed set if there are 2 roots, and an open set if there are a million or 0 roots. 8: If y varies immediately as x^2 we are in a position to assert y=x^2. If we double x, we get y=(2x)^2, that's y=4x^2. this means that y is prolonged via 4.

Answer 3

Consider the equation y = x² + 1. The graph has a vertex at (0,1) and opens upward, no negative values of y. So x² + 1 < 0 will have an empty solution set.

Answer 4

When it is never true.

For instance: solve x^2 < 0. There are no real solutions to this inequality.

Simlarly, no solutions to -x^2 > 0. Also no solutions to x^2+4x+4 < 0 since the polynomial is just (x+2)^2

Answer 5

when the question is incorrect!