The polynomial equation x^3-3 a x^2+left(27 a | vedclass

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Question

(a)

We have,

$x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$

Let

$f(x) =x^3-3 a x^2+\left(27 a^2+9\right) x+2016$

$f^{\prime}(x) =3 x^2-6

Vedclass · Accepted Answer

exactly one real root for any real $a$

Vedclass · Answer

(a)

We have,

$x^3-3 a x^2+\left(27 a^2+9ight) x+2016=0$

Let

$f(x) =x^3-3 a x^2+\left(27 a^2+9ight) x+2016$

$f^{\prime}(x) =3 x^2-6 a x+27 a^2+9$

$f^{\prime}(x) =3\left(x^2-2 a x+9 a^2+3ight)$

$f^{\prime}(x) =3\left((x-a)^2+8 a^2+3ight)$

$f^{\prime}(x) > 0, \forall x \in R$

$	herefore(x) 	ext { is increasing function, } \forall a \in R$.

$f(x)$ has exactly one real root for any real $\varepsilon$.

The polynomial equation $x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$ has

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