If alpha_1, alpha_2 and alpha_3 are the roots | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\alpha_1, \alpha_2, \alpha_3$ are the roots of $x^3+3x+2=0$.By Newton's Sums,let $S_n = \alpha_1^n + \alpha_2^n + \alpha_3^n$.The equa

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Given that $\alpha_1, \alpha_2, \alpha_3$ are the roots of $x^3+3x+2=0$.By Newton's Sums,let $S_n = \alpha_1^n + \alpha_2^n + \alpha_3^n$.The equation is $x^3 + 0x^2 + 3x + 2 = 0$.For $n=1$: $S_1 + 0 = 0 \Rightarrow S_1 = 0$.For $n=2$: $S_2 + 0(S_1) + 3(2) = 0 \Rightarrow S_2 = -6$.For $n=3$: $S_3 + 0(S_2) + 3(S_1) + 2(3) = 0 \Rightarrow S_3 = -6$.For $n=4$: $S_4 + 0(S_3) + 3(S_2) + 2(S_1) = 0$ $\Rightarrow S_4 + 3(-6) + 2(0) = 0$ $\Rightarrow S_4 = 18$.For $n=5$: $S_5 + 0(S_4) + 3(S_3) + 2(S_2) = 0 \Rightarrow S_5 + 3(-6) + 2(-6) = 0$.$S_5 - 18 - 12 = 0 \Rightarrow S_5 = 30$.

If $\alpha_1, \alpha_2$ and $\alpha_3$ are the roots of $x^3+3x+2=0$,then $\alpha_1^5+\alpha_2^5+\alpha_3^5=$

Similar Questions

If $x^2+2px-2p+8>0$ for all real values of $x$,then the set of all possible values of $p$ is

If $x$ is a real number,for what values of $x$ does $3x^2 + 14x + 11 > 0$ hold?

Let $S$ be the set of positive integral values of $a$ for which $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} < 0, \forall x \in R$. Then,the number of elements in $S$ is:

If $a$ and $b$ are the roots of the equation $x^2-7x-1=0$,then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to $........$.

The sum of the fourth powers of the roots of the equation $x^3+x+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module