How many real solutions does the equations $x^{7} + 14 x^{5} + 16 x^{3} + 30 x - 560 = 0$ have?.

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Solution

The correct option is A 1
$x^{7} + 14 x^{5} + 16 x^{3} + 30 x - 560 = 0$

Since, the no. of sign change in the sequence is one i.e., between $30 x$ &

$- 560$ . So, it has one positive root.
$p (- x) = - x^{7} - 14 x^{3} - 16 x^{3} - 30 x - 560 = 0$

Since there is no sign change in the sequence. So, there are no negative roots.
Total no. of real roots $= 1$ .

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