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Maths-

General

Easy

Question

How many real solutions does the equation $x to the power of 7 end exponent plus 14 x to the power of 5 end exponent plus 16 x to the power of 3 end exponent plus 30 x minus 560 equals 0$ have?

5
7
1
3

The correct answer is: 1

Let $f open parentheses x close parentheses equals x to the power of 7 end exponent plus 14 x to the power of 5 end exponent plus 16 x to the power of 3 end exponent plus 30 x minus 560$
$f to the power of ´ end exponent open parentheses x close parentheses equals 7 x to the power of 6 end exponent plus 70 x to the power of 4 end exponent plus 48 x to the power of 2 end exponent plus 30 greater than 0 comma blank for all x element of R$

$therefore f left parenthesis x right parenthesis$ is increasing

$therefore blank f open parentheses x close parentheses equals 0$ has only one solution

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