Exercise 3.6-Additional Problems - Chapter 3 - Theory of Equations - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Find the maximum possible number of real roots of the equation, $x^5-6 \cdot x^2-4 x+5=0$.
Solution:
Let $f(x)=x^5-6 x^2-4 x+5$
Check the terms when it changes sign
Number of changes $=2$
$\therefore$ Maximum number of positive real roots $=2$
$
\begin{aligned}
& f(-x)=(-x)^5-6(-x)^2-4(-x)+5 \\
& =-x^5-6 x^2+4 x+5
\end{aligned}
$
Check the terms when it changes signs.
Number of changes $=1$
$\therefore$ Maximum number of negative real roots $=1$
$\therefore$ Total maximum number of real roots $=2+1=3$

Question 2 .
Find the number of real roots of the equation, $x^2+5|x|+6=0$.
Solution:
$
\begin{array}{rr}
\left|\mathbf{x}^2\right|-5|\mathbf{x}|+6=0 & \\
(|\mathbf{x}|-2)(|\mathbf{x}|-3)=0 & \\
|x|-2=0 & |x|-3=0 \\
|x|=2 & |x|=3 \\
x=\pm 2 & x=\pm 3
\end{array}
$
It has four real roots. The real roots are $2,-2,3,-3$

Question 3.
Find the real roots of the equation, $x^2+5|x|+6=0$.
Solution:
$
x^2+5|x|+6=0
$
Case (i) If $x \geq 0$
$
\begin{aligned}
& x^2+5 x+6=0 \\
& (x+2)(x+3)=0 \\
& x=-2 \text { and } x=-3 \\
& \text { Case (ii) If } x<0 \\
& x^2-5 x+6=0 \\
& (x-2)(x-3)=0 \\
& x=2 \text { and } x=3
\end{aligned}
$
Question 4.
Solve $\mathrm{x}^4-4 \mathrm{x}^2+8 \mathrm{x}+35=0$. Given $(2+i \sqrt{3})$ is a root.
Solution:

Given $(2+i \sqrt{3})$ is a root of $\mathrm{P}(x)=0$
$\therefore(2-i \sqrt{3})$ is also a root of $\mathrm{P}(x)=0$
$\Rightarrow x^2-[(2+i \sqrt{3})+(2-i \sqrt{3})] x+[(2+i \sqrt{3})(2-i \sqrt{3})]$ is a factor of $\mathrm{P}(x)$
$\Rightarrow x^2-4 x+7$ is a factor of $\mathrm{P}(x)$. Dividing the polynomial $\mathrm{P}(x)=0$ by $x^2-4 x+7$.
$\Rightarrow x^2-4 x+7$ is a factor of $P(x)$. Dividing the polynomial $P(x)=0$ by $x^2-4 x+7$.
We get $x^2+4 x+5=0$ is a other factor. The roots of $x^2+4 x+5=0$ are
$
x=\frac{-4 \pm \sqrt{16-20}}{2}=-2 \pm i,
$
The roots are $-2 \pm i, 2 \pm i \sqrt{3}$

Question 5.
Solve $x^4-5 x^3+4 x^2+8 x-8=0$. Given $(1-\sqrt{5})$ is a root of the polynomial equation.
Solution:
Since $(1-\sqrt{5})$ is a root of the polynomial $\mathrm{P}(\mathrm{x})=0$
$(1+\sqrt{5})$ is also a root of $\mathrm{P}(\mathrm{x})=0$
$\Rightarrow \mathrm{x}^2-[(1+\sqrt{5})+(1-\sqrt{5})] \mathrm{x}+(1+\sqrt{5})(1-\sqrt{5})=0$ is a factor of $\mathrm{P}(\mathrm{x})=0 \Rightarrow \mathrm{x}^2-2 \mathrm{x}-4=$ 0 is a factor of $\mathrm{P}(\mathrm{x})=0$.
Dividing the polynomial by $x^2-2 x-4=0$
We get the other factor $x^2-3 x+2=0$
The roots of $x^2-3 x+2=0$
$(\mathrm{x}-2)(\mathrm{x}-1)=0$
$\mathrm{x}=1,2$
The roots are $1,2,1 \pm \sqrt{5}$
Question 6.
Find a polynomial equation of the lowest degree with rational co-efficient having $\sqrt{3},(1-2 i)$ as two of its roots.
Solution:
When $\sqrt{3}$ is a root, $-\sqrt{3}$ will also be a root.
Now the quadratic equation with $\sqrt{3},-\sqrt{3}$ are roots is $\mathrm{x}^2-(\sqrt{3}-\sqrt{3}) \mathrm{x}+(\sqrt{3})(-\sqrt{3})=0$
(i.e) $x^2-3=0$
When $1-2 \mathrm{i}$ is a root, $1+2 \mathrm{i}$ will be another root.
Now the quadratic equation with roots $1-2 \mathrm{i}$ and $1+2 \mathrm{i}$ is $\mathrm{x}^2-(1-2 \mathrm{i}+1+2 \mathrm{i}) \mathrm{x}+(1-2 \mathrm{i})(1+2 \mathrm{i})=0$
(i.e) $x^2-2 x+5=0$
$\therefore$ The equation with roots $\pm \sqrt{3}$ and $1 \pm 2 \mathrm{i}$ is $\left(\mathrm{x}^2-3\right)\left(\mathrm{x}^2-2 \mathrm{x}+5\right)=0$
(i.e) $x^4-2 x^3+5 x^2-3 x^2+6 x-15=0$,
(i.e) $x^4-2 x^3+2 x^2+6 x-15=0$