Exercise 3.1-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.
Construct a cubic equation with roots 2, 3, 4 .
Solution:
Given roots are 2, 3, 4
Take $\alpha=1 ; \beta=3 ; \gamma=4$
The required cubic polynomial is
$
\begin{aligned}
& x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma=0 \\
& x^3-(1+3+4) x^2+(3+12+4) x-12=0 \\
& x^3-8 x^2+19 x-12=0
\end{aligned}
$

Question 2.
If $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3-6 x^2+11 x-6=0$. From a cubic equation whose roots are $2 \alpha, 2 \beta, 2 \gamma$.
Solution:
Given that $\alpha, \beta, \gamma$ are the roots of $x^3-6 x^2+11 x-6=0 \ldots \alpha, \beta, \gamma$
$
\begin{aligned}
& \alpha+\beta+\gamma=6 \\
& \alpha \beta+\beta \gamma+\gamma \alpha=11 \\
& \alpha \beta \gamma=6 \ldots(3)
\end{aligned}
$
Form a cubic equation whose roots are $2 \alpha, 2 \beta, 2 \gamma$.
$
\begin{aligned}
& \therefore 2 \alpha+2 \beta+2 \gamma=2(\alpha+\beta+\gamma)=2(6)=12 \\
& 4 \alpha \beta+4 \beta \gamma+4 z \gamma \alpha=4(\alpha \beta+\beta \gamma+\gamma \alpha)=4(11)=44 \\
& (2 \alpha)(2 \beta)(2 \gamma)=8(\alpha \beta \gamma)=8(6)=48
\end{aligned}
$
The required cubic equation is
$
\begin{aligned}
& x^3-(2 \alpha+2 \beta+2 \gamma) x^2+(4 \alpha \beta+4 \beta \gamma+4 \gamma \alpha) x-(2 \alpha)(2 \beta)(2 \gamma)=0 \\
& x^3-12 x^2+44 x-48=0
\end{aligned}
$
Question 3.
If the roots of $x^4+5 x^3-30 x^2-40 x+64=0$ are in G.P; then find the roots.
Solution:
Let $a, a r, a r^2, a r^3$ be the roots of given equation.
$
x^4+5 x^3-30 x^2-40 x+64=0
$
Put $x=1$
$
\begin{aligned}
& 1+5-30-40+64=0 \quad \Rightarrow \therefore x=1 \text { is a root of the given equation. } \\
& \therefore \quad a=1 \quad \therefore \text { Product of the roots }=64 \\
& r^6=2^6 \quad \Rightarrow r=2 \\
&
\end{aligned}
$
The roots are $1,2,4,8$

Question 4.
Determine the value of $k$ such that the equation $(2 k-5) x^2-4 x-15=0$ and $(3 k-8) x^2-5 x-21=0$ may have a common root.
Solution:
If $\alpha$ be the common root, the two equations.
$(2 \mathrm{k}-5) \alpha^2-4 \alpha-15=0$
$(3 k-8) \alpha^2-5 \alpha-21=0$ These are the linear equation is $\alpha^2$ and $\alpha$.
By cross multiplication rule
$
\begin{aligned}
\frac{\alpha^2}{84-75} & =\frac{\alpha}{-15(3 k-8)+21(2 k-5)}=\frac{1}{-5(2 k-5)+4(3 k-8)} \\
\frac{\alpha^2}{9} & =\frac{\alpha}{-3 k+15}=\frac{1}{2 k-7} \\
\frac{\alpha^2}{9} & =\frac{\alpha}{3(5-k)}=\frac{1}{2 k-7}
\end{aligned}
$

Question 5.
If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+1=0$. Find the value of the following in terms of coefficients.
(i) $\sum \frac{1}{\beta r}$
(ii) $\sum \frac{1}{\alpha}$
(iii) $\sum \alpha^2 \beta$
Solution:
$\begin{aligned}
\alpha+\beta+\gamma & =-p \\
\alpha \beta+\beta \gamma+\gamma \alpha & =q \\
\alpha \beta \gamma & =-r
\end{aligned}$

(i)
$
\sum \frac{1}{\beta r}=\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+p+\gamma}{\alpha \beta \gamma}=\frac{p}{r}
$

(ii)
$
\sum \frac{1}{\alpha}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{-q}{r}
$

(iii)
$
\begin{aligned}
\sum \alpha^2 \beta & =\alpha^2 \beta+\beta^2 \alpha+\gamma^2 \alpha+\alpha^2 \gamma+\beta^2 \gamma+\gamma^2 \beta \\
& =(\alpha \beta+\beta \gamma+\alpha \gamma)(\alpha+\beta+\gamma)-3 \alpha \beta \gamma \\
& =(q \times p)-3(-r)=-p q+3 r=3 r-p q
\end{aligned}
$

Question 6.
$\alpha, \beta, \gamma$ are the roots of the equation $x^3+3 x^2+2 x+1=0$. Find $\sum \alpha^3$ and $\sum \alpha^{-2}$.
Solution:
$
\begin{aligned}
\alpha+\beta+\gamma & =-3 \\
\alpha \beta+\beta \gamma+\gamma \alpha & =2 \\
\alpha \beta \gamma & =-1 \\
\left(a^3+b^3+c^3\right)-3 a b c & =\left(a^2+b^2+c^2-a b-b c-c a\right)(a+b+c) \\
\sum \alpha^3 & =\left(\alpha^3+\beta^3+\gamma^3\right)=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)+3 \alpha \beta \gamma \\
& =(\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2-2 \alpha \beta-2 \beta \gamma-2 \gamma \alpha-\alpha \beta-\beta \gamma-\gamma \alpha\right]+3 \alpha \beta \gamma \\
& =(\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2-3(\alpha \beta-\beta \gamma-\gamma \alpha)\right]+3 \alpha \beta \gamma \\
& =(-3)[9-3(2)]+3(-1)=(-3)(9-6)-3=(-3)(3)-3=-9-3=-12 \\
\sum \alpha^{-2} & =\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2} \\
& =\frac{(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2\left[\alpha \gamma \beta^2+\alpha \beta \gamma^2+\beta \gamma \alpha^2\right]}{(\alpha \beta \gamma)^2} \\
& =\frac{(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2 \alpha \gamma \beta+[\beta+\gamma+\alpha]}{(\alpha \beta \gamma)^2}=\frac{(2)^2-2(-1)(-3)}{1}=4-6=-2
\end{aligned}
$