Exercise 3.6 - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.6$
Question 1.
Discuss the maximum possible number of positive and negative roots of the polynomial equation $9 x^9-4 x^8+4 x^7-3 x^6+2 x^5+x^3+7 x^2+7 x+2=0$
Solution:
$
P(x)=9 x^9-4 x^8+4 x^7-3 x^6+2 x^5+x^3+7 x^2+7 x+2
$
The number of sign changes in $\mathrm{P}(\mathrm{x})$ is 4 .
$\therefore \mathrm{P}(\mathrm{x})$ has atmost 4 positive roots.
$
P(-x)=-9 x^9-4 x^8-4 x^7-3 x^6-2 x^5-x^3+7 x^2-7 x+2
$
The number of sign changes in $\mathrm{P}(-\mathrm{x})$ is 3 .
$\mathrm{P}(\mathrm{x})$ has almost 3 negative roots. Since the difference between the number of sign changes in coefficient $\mathrm{P}(-\mathrm{x})$ and the number of negative roots of the polynomial $\mathrm{P}(\mathrm{x})$ is even.
The number of negative roots $=$ at most 2 .
Question 2.
Discuss the maximum possible number of positive and negative zeros of the polynomials $x^2-5 x$ $+6$ and $x^2-5 x+16$. Also, draw a rough sketch of the graphs.
Solution:
$
P(x)=x^2-5 x+6
$
The number of sign changes in $P(x)$ is 2 .
$\mathrm{P}(\mathrm{x})$ has atmost 2 positive roots. $\mathrm{P}(-\mathrm{x})=\mathrm{x}^2+5 \mathrm{x}+6$.
The number of sign changes in $P(-x)$ is 0 .
$\therefore \mathrm{P}(\mathrm{x})$ has no negative roots. $\mathrm{P}(\mathrm{x})=\mathrm{x}^2-5 \mathrm{x}+16$

Question 3.
Show that the equation $\mathrm{x}^9-5 \mathrm{x}^5+4 \mathrm{x}^4+2 \mathrm{x}^2+1=0$ has atleast 6 imaginary solutions.
Solution:
$
P(x)=x^9-5 x^5+4 x^4+2 x^2+1
$
(i) The number of sign changes in $\mathrm{P}(\mathrm{x})$ is 2 . The number of positive roots is atmost 2 .
(ii) $P(-x)=-x^9+5 x^5+4 x^4+2 x^2+1$. The number of sign changes in $P(-x)$ is 1 . The number of negative roots of $P(x)$ is atmost 1 . Since the difference of number of sign changes in $P(-x)$ and number of negative zeros is even.
$\mathrm{P}(\mathrm{x})$ has one negative root.
(iii) 0 is not the zero of the polynomial $\mathrm{P}(\mathrm{x})$. So the number of real roots is almost 3 .
$\therefore$ The number of imaginary roots at least 6 .
Question 4.
Determine the number of positive and negative roots of the equation $x^9-5 x^8-14 x^7=0$.
Solution:
$
x^9-5 x^8-14 x^7=0
$
$P(x)=x^9-5 x^8-14 x^7$. The number of sign changes is $P(x)$ is 1 .
The number of positive roots is 1 . $P(-x)=-x^9-5 x^8+14 x^7$
The number of sign changes is $\mathrm{P}(-\mathrm{x})$ is one. The number of negative zero of $\mathrm{P}(-\mathrm{x})$ is 1 . It is clear that 0 is a root of the equation.
$\therefore$ The number of the imaginary roots is at least 6 .
Question 5 .
Find the exact number of real zeros and imaginary of the polynomial $x^9+9 x^7+7 x^5+5 x^3+3 x$.
Solution:
$
\mathrm{P}(\mathrm{x})=\mathrm{x}^9+9 \mathrm{x}^7+7 \mathrm{x}^5+5 \mathrm{x}^3+3 \mathrm{x} \text {. }
$
There is no change in the sign of $\mathrm{P}(\mathrm{x})$ and $\mathrm{P}(-\mathrm{x}), \mathrm{P}(\mathrm{x})$ has no positive and no negative real roots, but 0 is the root of the polynomial equation $\mathrm{P}(\mathrm{x})$.