One root of following given equation 2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0 is

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4-2.Quadratic Equations and Inequations

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One root of the following given equation $2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0$ is

A

$1$

B

$3$

C

$5$

D

$7$

Solution

(c) Putting $x = 5$

$2{(5)^5} – 14{(5)^4} + 31{(5)^3} – 64{(5)^2} + 19(5) + 130 = 0$

Hence $x = 5$ satisfies the given equation.

Thus $5$ is a root of the equation.

Standard 11

Mathematics

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