Exercise 3.2 - 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 $3.2$
Question 1.
If $\mathrm{k}$ is real, discuss the nature of the roots of the polynomial equation $2 \mathrm{x}^2+\mathrm{kx}+\mathrm{k}=0$, in terms of $k$.
Solution:
The given quadratic equation is $2 \mathrm{x}^2+\mathrm{kx}+\mathrm{k}=0$
$
\begin{aligned}
& \mathrm{a}=2, \mathrm{~b}=\mathrm{k}, \mathrm{c}=\mathrm{k} \\
& \Delta=\mathrm{b}^2-4 \mathrm{ac}=\mathrm{k}^2-4(2) \mathrm{k}=\mathrm{k}^2-8 \mathrm{k}
\end{aligned}
$
(i) If the roots are equal
$
\begin{aligned}
& \mathrm{k}^2-8 \mathrm{k}=0 \\
& \Rightarrow \mathrm{k}(\mathrm{k}-8)=0 \\
& \Rightarrow \mathrm{k}=0, \mathrm{k}=8
\end{aligned}
$
(ii) If the roots are real
$
\begin{aligned}
& \mathrm{k}^2-8 \mathrm{k}>0 \\
& \mathrm{k}(\mathrm{k}-8)>0 \\
& \mathrm{k} \in(-\infty, 0) \cup(8, \infty)
\end{aligned}
$
(iii) If this roots are imaginary
$
\begin{aligned}
& \mathrm{k}^2-8 \mathrm{k}<0 \\
& \Rightarrow \mathrm{k} \in(0,8)
\end{aligned}
$
Question 2.
Find a polynomial equation of minimum degree with rational coefficients, having $2+\sqrt{3}$ i as a root.
Solution:
Given roots is $(2+\sqrt{ } 3$ i)
The other root is $(2-\sqrt{3} \mathrm{i})$, since the imaginary roots with real co-efficient occur as conjugate pairs.
$
\begin{aligned}
& x^2-x(\text { S.O.R })+\text { P.O.R }=0 \\
& \Rightarrow x^2-x(4)+(4+3)=0 \\
& \Rightarrow x^2-4 x+7=0
\end{aligned}
$

Question 3.
Find a polynomial equation of minimum degree with rational coefficients, having $2 \mathrm{i}+3$ as a root.
Solution:
Given roots is $(3+2 \mathrm{i})$, the other root is $(3-2 \mathrm{i})$;
Since imaginary roots occur in with real co-efficient occurring conjugate pairs.
$
\begin{aligned}
& x^2-x(\text { S.O.R })+\text { P.O.R }=0 \\
& \Rightarrow x^2-x(6)+(9+4)=0 \\
& \Rightarrow x^2-6 x+13=0
\end{aligned}
$
Question 4.
Find a polynomial equation of minimum degree with rational coefficients, having $\sqrt{5}-\sqrt{3}$ as a root.
Solution:
The given one roots of the polynomial equation are $(\sqrt{ } 5-\sqrt{3})$
The other roots are $(\sqrt{5}+\sqrt{3}),(-\sqrt{5}+\sqrt{3})$ and $(-\sqrt{5}-\sqrt{3})$.
The quadratic factor with roots $(\sqrt{5}-\sqrt{ } 3)$ and $(\sqrt{5}+\sqrt{3})$ is
$
\begin{aligned}
& =x^2-x(\text { S.O.R })+\text { P.O.R } \\
& =x^2-x(2 \sqrt{5})+(\sqrt{5}-\sqrt{ } 3)(\sqrt{5}+\sqrt{ } 3) \\
& =x^2-2 \sqrt{5} x+2
\end{aligned}
$
The other quadratic factors with roots $(-\sqrt{ } 5+\sqrt{3})(-\sqrt{ } 5-\sqrt{3})$ is
$
\begin{aligned}
& =x^2-x(\text { S.O.R })+\text { P.O.R } \\
& =x^2-x(-2 \sqrt{5})+(5-3) \\
& =x^2+2 \sqrt{ } 5 x+2
\end{aligned}
$
To rationalize the co-efficients with minimum degree
$
\begin{aligned}
& \left(x^2-2 \sqrt{5} \mathrm{x}+2\right)\left(\mathrm{x}^2+2 \sqrt{5} \mathrm{x}+2\right)=0 \\
& \Rightarrow\left(\mathrm{x}^2+2\right)^2-(2 \sqrt{5} \mathrm{x})^2=0 \\
& \Rightarrow \mathrm{x}^4+4+4 \mathrm{x}^2-20 \mathrm{x}^2=0 \\
& \Rightarrow \mathrm{x}^4-16 \mathrm{x}^2+4=0
\end{aligned}
$
Question 5.
Prove that a straight line and parabola cannot intersect at more than two points.
Solution:
Let the standard equation of parabola $y^2=4 a x$
Equation of line be $\mathrm{y}=\mathrm{mx}+\mathrm{c}$...(2)
Solving (1) \& (2)
$
\begin{aligned}
& (m x+c)^2=4 a x \\
& \Rightarrow m x^2+2 m c x+c^2-4 a x=0
\end{aligned}
$

$
\Rightarrow m x^2+2 x(m c-2 a)+c^2=0
$
This equation can not have more than two solutions and hence a line and parabola cannot intersect at more than two points.