Exercise 3.3-Additional Problems - Chapter 3 - Theory of Equations - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
If one root of $x^3+2 x^2+3 x+k=0$ is sum of the other two roots then find the value of $k$.

Solution:
Let $\alpha, \beta, \gamma$ be the roots of given equation.
But, $\alpha=\beta+\gamma$ $\alpha+\beta+\gamma=-2 \ldots(1) \Rightarrow 2 \alpha=-2 \Rightarrow \alpha=-1$
$\alpha \beta+\beta \gamma+\gamma \alpha=3 \ldots$ (2) This gives $\beta+\gamma=-1$
$
\begin{aligned}
& \alpha \beta \gamma=-\mathrm{k} \text {...(3) } \\
& (2) \Rightarrow(-1) \beta+\beta(-1-\beta)+(-1-\beta)+(-1)=3 \\
& -\beta-\beta-\beta^2+1+\beta=3 \\
& -\beta-\beta^2=2 \\
& \text { (3) } \Rightarrow \\
& k=-\alpha \beta \gamma \\
& k=-(-1) \beta \gamma \quad \Rightarrow k=\beta \gamma \\
& k=\beta(-1-\beta)=-\beta-\beta^2 \\
& k=2 \\
&
\end{aligned}
$

Question 2.
If sum of the roots of the equation $x^3-3 x^2-16 x+k=0$ is zero then find the value of $k$.
Solution:
Let $\alpha, \beta, \gamma$ be the roots of given equation.
But, $\alpha+\beta=0$
But,
$
\begin{aligned}
& \alpha+\beta+\gamma=3 \\
& \alpha \beta+\beta \gamma+\gamma \alpha=-16 \\
& \alpha \beta \gamma=-k \\
& (1) \Rightarrow \quad \gamma=3 \\
& \text { (3) } \Rightarrow \quad \alpha \beta=\frac{-k}{3} \\
& \text { (2) } \Rightarrow \frac{-k}{3}+3 \beta+3 \alpha=-16 \quad \Rightarrow \frac{-k}{3}+3(\beta+\alpha)=-16 \quad(\text { Since } \alpha+\beta=0) \\
& \Rightarrow \quad \frac{-k}{3}=-16 \quad \Rightarrow k=48 \\
&
\end{aligned}
$
Question 3.
Find all zeros of the polynomial $x^3-5 x^2+9 x-5=0$, If $2+i$ is a root.

Solution:
Given root is $2+i$.
Other root is $2-i$
S.R: $\quad 2+i+2-i=4$
P.R: $\quad(2+i)(2-i)=4+1=5$
$\therefore$ The factor : $x^2-x($ S.R.) $+(\mathrm{PR})$.
$
: x^2-4 x+5
$

$(x-1)$ is the other factor.
$\therefore$ The roots are $2+i, 2-i, 1$