Show that the equation $2 x^{7} - x^{4} + 4 x^{3} - 5 = 0$ has at least four imaginary roots.

Open in App

Solution

Give equation, $2 x^{7} - x^{4} + 4 x^{3} - 5 = 0$ , has three changes of signs. Hence there can be a maximum of 3 positive roots.

$f (- x) = - 2 x^{7} - x^{4} - 4 x - 5 = 0$ , which has no changes of signs. Hence the given equation has zero negative roots.

Now, as the equation is of $7^{t h}$ degree, it must have at least $(7 - 3) = 4$ imaginary roots.

Suggest Corrections

Similar questions

x^{4} - 4 x^{3} + a x^{2} + b x + 1 = 0

x^{9} - x^{5} + x^{4} + x^{2} + 1 = 0

x^{4} + 4 x^{3} - 2 x^{2} - 12 x + 9 = 0

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0

\sqrt{- 1}

x^{2} + b_{1} x + c_{1} = 0

x^{2} + b_{2} x + c_{2} = 0

b_{1} b_{2} = 2 (c_{1} + c_{2})

b_{1}, b_{2}, c_{1}, c_{2} \in R,

Join BYJU'S Learning Program