(C) To find the root,we substitute the given options into the equation $f(x) = 2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$.For $x = 5$:$f(5) = 2(5)^

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(C) To find the root,we substitute the given options into the equation $f(x) = 2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$.For $x = 5$:$f(5) = 2(5)^5 - 14(5)^4 + 31(5)^3 - 64(5)^2 + 19(5) + 130$$f(5) = 2(3125) - 14(625) + 31(125) - 64(25) + 95 + 130$$f(5) = 6250 - 8750 + 3875 - 1600 + 95 + 130$$f(5) = (6250 + 3875 + 95 + 130) - (8750 + 1600)$$f(5) = 10350 - 10350 = 0$Since $f(5) = 0$,$x = 5$ is a root of the equation.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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

A
$1$
B
$3$
C
$5$
D
$7$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

If $f(x)$ is a second degree polynomial such that $f(x) \geq 0$ for all $x \in R$,$f(-3) = 0$,and $f(0) = 18$,then find $f(3)$.

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If one root of the quadratic equation $x^2 + px + (1 - p) = 0$ is $(1 - p)$,then what are its roots?

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Let $p(x)$ be a quadratic polynomial such that $p(0) = 1$. If $p(x)$ leaves a remainder of $4$ when divided by $x - 1$ and a remainder of $6$ when divided by $x + 1$,then:

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If $\alpha$ and $\alpha^2$ are the roots of the equation $x^2 + x + 1 = 0$,then the equation whose roots are $\alpha^{31}$ and $\alpha^{62}$ is .....

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The number of pairs $(a, b)$ of real numbers such that whenever $\alpha$ is a root of the equation $x^{2}+ax+b=0$,$\alpha^{2}-2$ is also a root of this equation,is:

DifficultJEE MAIN 2021

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

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If $f(x)$ is a second degree polynomial such that $f(x) \geq 0$ for all $x \in R$,$f(-3) = 0$,and $f(0) = 18$,then find $f(3)$.

If one root of the quadratic equation $x^2 + px + (1 - p) = 0$ is $(1 - p)$,then what are its roots?

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