Question

of s1 s2 s3........sr are the sum of products of the roots taken r at a time then x^5-x^2+4x-9=0 then s3+s4-s5=​

Answer 1

summatin of first 5 G.P series, S5=1023

and common ratio, r=4

summation of G.P terms having n number of terms =\frac{a\times(r^n -1)}{r-1}

r−1

a×(r

n

−1)

hence, summation of first five terms = \frac{a\times(r^5 -1)}{r-1}

r−1

a×(r

5

−1)

Answer 2

The value of s3+s4-s5 =​ - 4.

Step-by-step explanation:

Given:

The equation $x^{5} -x^{2} +4x-9=0$

To Find:

The value of s3+s4-s5.

Formula Used:

The equation $ax^{5} + bx^{4} + cx^{3} +dx^{2} + e x+f=0$.    

 then,

 $s1 =\frac{ cofficient\ of x^{4} }{ cofficient \ of \ x^{5} } = - \frac{b}{a}$

  $s2 =\frac{ cofficient\ of x^{3} }{ cofficient \ of \ x^{5} } = \frac{c}{a}$

   $s3 =\frac{ cofficient\ of x^{2} }{ cofficient \ of \ x^{5} } = - \frac{d}{a}$

  $s4=\frac{ cofficient\ of x^{} }{ cofficient \ of \ x^{5} } = \frac{e}{a}$

   $s5=\frac{ cofficient\ of constant }{ cofficient \ of \ x^{5} } = - \frac{f}{a}$

Solution:

As given- the equation  $x^{5} -x^{2} +4x-9=0$.

Comparing cofficients of equation $x^{5} -x^{2} +4x-9=0$ with equation $ax^{5} + bx^{4} + cx^{3} +dx^{2} + e x+f=0$ , we get.

a= 1,    b=0,     c=0,    d= -1 ,  e =4     and  f = -9

$s3 =\frac{ cofficient\ of x^{2} }{ cofficient \ of \ x^{5} } = - \frac{d}{a}$

                               $= - \frac{(-1)}{1} = 1$

$s4=\frac{ cofficient\ of x^{} }{ cofficient \ of \ x^{5} } = \frac{e}{a}$

                                 $=\frac{4}{1} =4$

$s5=\frac{ cofficient\ of constant }{ cofficient \ of \ x^{5} } = - \frac{f}{a}$

                                      $= - \frac{(-9)}{1} = 9$

Therefore  s3+s4-s5 = 1+4-9 = - 4

                                 

Thus, the value of s3+s4-s5=​ - 4.

Correct Question

If s1 s2 s3........sr is the sum of products of the roots taken r at a time then x^5-x^2+4x-9=0 then s3+s4-s5=​.