Exercise 3.2-Additional Problems - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Find a polynomial equation of minimum degree with rational co-efficients having $1-i$ as a root.
Solution:
Given root is $1-\mathrm{i}$
The other root is $1+\mathrm{i}$
Sum of the roots: $1-\mathrm{i}+1+\mathrm{i}=2$
product of the roots: $(1-\mathrm{i})(1+\mathrm{i})=(1)^2+(1)^2 \Rightarrow 1+1=2$
$\therefore$ The required polynomial equation of minimum degree with rational coefficients is $\mathrm{x}^2-\mathrm{x}$ (S.R.) $)+($ P.R. $)=0$ $x^2-2 x+2=0$
Question 2.
Find a polynomial equation of minimum degree with rational co-efficients having $\sqrt{3}+\sqrt{7}$ as a root.
Solution:
Given root is $\sqrt{3}+\sqrt{7}$
The other root is $\sqrt{3}-\sqrt{7}$
S.R: $\sqrt{3}+\sqrt{7}+\sqrt{3}-\sqrt{7}=2 \sqrt{3}$
P.R: $(\sqrt{3}+\sqrt{7})(\sqrt{3}-\sqrt{7})=(\sqrt{3})^2-(\sqrt{7})^2=3-7=-4$
One of the factor is : $x^2-x(\mathrm{~S} . \mathrm{R})+(\mathrm{P} . \mathrm{R})$
$
: x^2-2 \sqrt{3} x-4
$
The other factor also will be : $x^2+2 \sqrt{3} x-4$
The required polynomial equation of minimum degree
$
\begin{aligned}
\left(x^2-2 \sqrt{3} x-4\right)\left(x^2+2 \sqrt{3} x-4\right)=0 & \\
\left(x^2-4-2 \sqrt{3} x\right)\left(x^2-4+2 \sqrt{3} x\right)=0 & \Rightarrow\left(x^2-4\right)^2-(2 \sqrt{3} x)^2=0 \\
x^4+16-8 x^2-12 x^2=0 & \Rightarrow x^4-20 x^2+16=0
\end{aligned}
$

Question 3.
If the roots of the equation $x^3+p x^2+q x+r=0$ are in A.P then show that $2 p^3-9 p q+27 r=0$.

Solution:
Let the roots of the given equation is $\mathrm{a}-\mathrm{d}, \mathrm{a}, \mathrm{a}+\mathrm{d}$
$
\begin{aligned}
s_1=a-d+a+a+d= & 3 a \\
\text { (i.e) } 3 a=-p & \Rightarrow a=\frac{-p}{3}
\end{aligned}
$
Since $a$ is a root of $p(x)=0$
$
\begin{array}{cl}
\left(\frac{-p}{3}\right)^3+p\left(\frac{-p}{3}\right)^2+q\left(\frac{-p}{3}\right)+r=0 & \\
\frac{-p^3}{27}+\frac{p^3}{9}-\frac{p q}{3}+r=0 & \Rightarrow \frac{2 p^3}{27}-\frac{9 p q}{27}+\frac{27 r}{27}=0 \\
2 p^3-9 p q+27 r=0 &
\end{array}
$
Question 4.
Solve $27 x^3+42 x^2-28 x-8=0$ given that its roots are in geometric progressive.
Solution:
Let the roots be $\frac{a}{r}, a, a r$
$
\begin{aligned}
\frac{a}{r} \cdot a \cdot a r & =\frac{8}{27} \\
a^3 & =\frac{2^3}{3^3} \quad \Rightarrow a=\frac{2}{3}
\end{aligned}
$
$a=\frac{2}{3}$ is a root, $(x-2 / 3)$ is a factor on division the other factor of the polynomial is $27 x^2+60 x+12$
The roots are
$
\therefore \quad x=\frac{-2}{9} \text { or }-2 \quad \therefore \text { The roots are } \frac{-2}{9},-2,-2
$

Question 5 .
Solve the equation $15 x^3-23 x^2+9 x-1=0$. Where roots are in H.P.
Solution:
[If $a, b, c$ are in H.P. then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. the we have $b=\frac{2 a c}{a+c}$ ]
Let $\alpha, \beta, \gamma$ be the roots of the equation
$
\begin{array}{r}
\alpha \beta+\beta \gamma+\gamma \alpha=\frac{9}{15} \\
\alpha \beta \gamma=\frac{1}{15}
\end{array}
$
Since $\alpha, \beta, \gamma$ are in H.P
$
\begin{aligned}
& \beta=\frac{2 \alpha \gamma}{\alpha+\gamma} \\
& \alpha \beta+\beta \gamma=2 \alpha \gamma \\
& \therefore(1) \Rightarrow \quad 3 \alpha \gamma=\frac{9}{15} \quad \Rightarrow \alpha \gamma=\frac{3}{15}=\frac{1}{5} \\
& \therefore(2) \Rightarrow \quad(\alpha \gamma) \beta=\frac{1}{15} \\
& \frac{1}{5} \beta=\frac{1}{15} \quad \Rightarrow \beta=\frac{1}{3} \\
&
\end{aligned}
$
$\beta=\frac{1}{3}$ is a roots of the given polynomial proceeding as on the above problem.
$\therefore$ The roots are $\frac{1}{3}, 1, \frac{1}{5}$