Exercise 3.1 - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Theory of Equations
Ex $3.1$
Question 1.
If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.
Solution:
Let the side of the cube be ' $x$ '
Sides of cuboid are $(x+1)(x+2)(x+3)$
$\therefore$ Volume of cuboid $=\mathrm{x}^3+52$
$\Rightarrow(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)=\mathrm{x}^3+52$
$\Rightarrow\left(\mathrm{x}^2+3 \mathrm{x}+2\right)(\mathrm{x}+3)=\mathrm{x}^3+52$
$\Rightarrow \mathrm{x}^3+3 \mathrm{x}^2+3 \mathrm{x}^2+9 \mathrm{x}+2 \mathrm{x}+6-\mathrm{x}^3-52=0$
$\Rightarrow 6 \mathrm{x}^2+11 \mathrm{x}-46=0(\div 2)$
$\Rightarrow(\mathrm{x}-2)(6 \mathrm{x}+23)=0$
$\Rightarrow \mathrm{x}-2=0$ or $6 \mathrm{x}+23=0$
$\Rightarrow \mathrm{x}=2$ or $\mathrm{x}=-\frac{23}{6}$ (not possible)
$\therefore \mathrm{x}=2$
Volume of cube $=2^3=8$
Volume of cuboid $=52+8=60$ cubic units

Question 2.

Construct a cubic equation with roots (i) 1,2 and 3
(ii) 1,1 and $-2$
(iii) 2, and 1 $\frac{1}{2}$
Solution:
(i) Given roots are $\alpha=1, \beta=2, \gamma=3$
The cubic equation is
$
\begin{aligned}
& x^3-x^2(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma=0 \\
& \Rightarrow x^3-x^2(1+2+3)+x(2+6+3)-(1)(2)(3)=0 \\
& \Rightarrow x^3-6 x^2+11 x-6=0
\end{aligned}
$
(ii) $\alpha=1, \beta=1, \gamma=-2$
The cubic equation is
$
\begin{aligned}
& x^3-x^2(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma=0 \\
& \Rightarrow x^3-x^2(1+1-2)+x(1-2-2)-(1)(1)(-2)=0 \\
& \Rightarrow x^3-0 x^2-3 x+2=0
\end{aligned}
$

$
\Rightarrow \mathrm{x}^3-3 \mathrm{x}+2=0
$
(iii) $\alpha=2, \beta=\frac{1}{2}, \gamma=1$
The cubic equation is
$
\begin{aligned}
& x^3-x^2(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)-\alpha \beta \gamma=0 \\
& x^3-x^2\left(2+\frac{1}{2}+1\right)+x\left(1+\frac{1}{2}+2\right)-(2)\left(\frac{1}{2}\right)(1)=0 \\
& x^3-x^2\left(\frac{4+1+2}{2}\right)+x\left(\frac{2+1+4}{2}\right)-1=0 \\
& x^3-x^2\left(\frac{7}{2}\right)+x\left(\frac{7}{2}\right)-1=0 \\
& 2 x^3-7 x^2+7 \mathrm{x}-2=0
\end{aligned}
$

Question 3.

If $\alpha, \beta$ and $\gamma$ are the roots of the cubic equation $\mathrm{x}$ ${ }^3+2 x^2+3 x+4=0$, form a cubic equation whose roots are
(i) $2 \alpha, 2 \beta, 2 \gamma$
(ii) $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$
(iii) $-\alpha,-\beta,-\gamma$
Solution:
(i) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+2 x^2+3 x+4=0$
Compare with $x^3+b x^2+c x+d=0$
$
\begin{aligned}
& \mathrm{b}=2, \mathrm{c}=3, \mathrm{~d}=4 \\
& \alpha+\beta+\gamma=-6=-2 \\
& \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{c}=3 \\
& \alpha \beta \gamma=-d=-4
\end{aligned}
$
Given roots are $2 \alpha, 2 \beta, 2 \gamma$
$
\begin{aligned}
& 2 \alpha+2 \beta+2 \gamma=2(\alpha+\beta+\gamma) \\
& =2(-2) \\
& =-4 \\
& (2 \alpha)(2 \beta)+(2 \beta)(2 \gamma)+(2 \gamma)(2 \alpha)=(4 \alpha \beta+4 \beta \gamma+4 \gamma \alpha) \\
& =4(\alpha \beta+\beta \gamma+\gamma \alpha) \\
& =4(3) \\
& =12 \\
& (2 \alpha)(2 \beta)(2 \gamma)=8(\alpha \beta \gamma) \\
& =8(-4) \\
& =-32
\end{aligned}
$
The equation is
$
\begin{aligned}
& x^3-x^2(2 \alpha+2 \beta+2 \gamma)+x(4 \alpha \beta+4 \beta \gamma+4 \gamma \alpha)-8(\alpha \beta \gamma)=0 \\
& \Rightarrow x^3-x^2(-4)+x(12)-(-32)=0 \\
& \Rightarrow x^3+4 x^2+12 x+32=0
\end{aligned}
$
(ii) The given roots are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$
The cubic equation is

\begin{aligned}
x^3-x^2\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)+x\left(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\right)-\frac{1}{\alpha \beta \gamma} & =0 \\
x^3-x^2\left(\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}\right)+x\left(\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}\right)-\frac{1}{\alpha \beta \gamma} & =0 \\
x^3-x^2\left(\frac{3}{-4}\right)+x\left(\frac{-2}{-4}\right)-\left(\frac{1}{-4}\right) & =0 \\
x^3+\frac{3 x^2}{4}+\frac{2 x}{4}+\frac{1}{4} & =0
\end{aligned}
$
$4 \mathrm{x}^3+3 \mathrm{x}^2+2 \mathrm{x}+1=0$ (Multiply by 4$)$
(iii) The given roots are $-\alpha,-\beta,-\gamma$
The cubic equation is
$
\begin{aligned}
& x^3-x^2(-\alpha-\beta-\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)+(\alpha \beta \gamma)=0 \\
& \Rightarrow x^3+x^2(\alpha+\beta+\gamma)+x(\alpha \beta+\beta \gamma+\gamma \alpha)+(\alpha \beta \gamma)=0 \\
& \Rightarrow x^3+x^2(-2)+x(3)-4=0 \\
& \Rightarrow x^3-2 x^2+3 x-4=0
\end{aligned}
$

Question 4.
Solve the equation $3 x^3-16 x^2+23 x-6=0$ if the product of two roots is 1 .
Solution:
The given equation is $3 \mathrm{x}^3-16 \mathrm{x}^2+23 \mathrm{x}-6=0$
$
\Rightarrow x^3-\frac{16}{3} x^2+\frac{23}{3} x-2=0 \quad(\div 3)
$
Let the roots be $\alpha, \beta, \gamma$
$
\begin{aligned}
& \alpha+\beta+\gamma=-\mathrm{b}=\frac{16}{3} \ldots \\
& \alpha \beta+\beta \gamma+\gamma \alpha=\mathrm{c}=\frac{23}{3} \\
& \alpha \beta \gamma=-\mathrm{d}=2 \ldots . .(3)
\end{aligned}
$
Given that $\alpha \beta=1$
from (3), $\gamma=2$
Substitute $\beta=\frac{1}{\alpha}, \gamma=2$ in (1)
$
\Rightarrow \alpha+\frac{1}{\alpha}+2=\frac{16}{3}
$
$
\begin{aligned}
& \Rightarrow \frac{\alpha^2+1}{\alpha}=\frac{16}{3}-2 \\
& \Rightarrow \frac{\alpha^2+1}{\alpha}=\frac{10}{3} \\
& \Rightarrow 3 \alpha^2+3=10 \alpha \\
& \Rightarrow 3 \alpha^2-10 \alpha+3=0 \\
& \Rightarrow(3 \alpha-1)(\alpha-3)=0
\end{aligned}
$
$\Rightarrow \alpha=\frac{1}{3}, 3$
$\alpha=\frac{1}{3}, \beta=3$ (or) when $\mathrm{a}=3, \beta=\frac{1}{3}$
$\therefore$ The roots are $3, \frac{1}{3}, 2$
(or) when $\gamma=2$, by synthetic division method.

Question 5.
Find the sum of squares of roots of the equation $2 x^4-8 x^3+6 x^2-3=0$.
Solution:
The given equation is $2 x^4-8 x^3+6 x^2-3=0$.
$(\div 2) \Rightarrow x^4-4 x^3+3 x^2-\frac{3}{2}=0$
Let the roots be $\alpha, \beta, \gamma, \delta$
$
\begin{aligned}
& \alpha+\beta+\gamma+\delta=-b=4 \\
& (\alpha \beta+\beta \gamma+\gamma \delta+\alpha \delta+\alpha \gamma+\beta \delta)=c=3 \\
& \alpha \beta \gamma+\beta \gamma \delta+\gamma \delta \alpha=-d=0 \\
& \alpha \beta \gamma \delta=\frac{-3}{2} \\
& \text { To Find } \alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha \beta+\beta \gamma+\gamma \delta+\alpha \delta+\alpha \gamma+\beta \delta) \\
& =(4)^2-2(3) \\
& =16-6 \\
& =10
\end{aligned}
$

Question 6.

Solve the equation $x^3-9 x^2+14 x+24=0$ if it is given that two of its roots are in the ratio $3: 2$.
Solution:
The given equation is $x^3-9 x^2+14 x+24=0$.
Since the two roots are in the ratio $3: 2$.
The roots are $\alpha, 3 \lambda, 2 \lambda$

$\begin{aligned}
& \alpha+3 \lambda+2 \lambda=-b=9 \\
& \Rightarrow \alpha+5 \lambda=9 \ldots \ldots(1) \\
& (\alpha)(3 \lambda)(2 \lambda)=-24 \\
& 6 \lambda^2 \alpha=-24 \\
& \Rightarrow \lambda^2 \alpha=-4 \ldots \ldots(2) \\
& (1) \Rightarrow \alpha=9-5 \lambda \\
& (2) \Rightarrow \lambda^2(9-5 \lambda)=-4 \\
& 9 \lambda^2-5 \lambda^3+4=0 \\
& 5 \lambda^3-9 \lambda^2-4=0
\end{aligned}$

$
(\lambda-2)\left(5 \lambda^2+\lambda+2\right)=0
$
$\lambda=2,5 \lambda^2+\lambda+2=0$ has only Imaginary roots $\Delta<0$ when $\lambda=2, \alpha=9-5(2)=9-10=-1$
The roots are $\alpha, 3 \lambda, 2 \lambda$ i.e., $-1,6,4$

Question 7.
If $\alpha, \beta$ and $\gamma$ are the roots of the polynomial equation $\mathrm{ax}^3+b x^2+c x+d=0$, find the value of $\Sigma \frac{\alpha}{\beta \gamma}$ in terms of the coefficients.
Solution:
The given equation is $a x^3+b x^2+c x+d=0$.
$
\div \mathrm{a} \Rightarrow x^3+\frac{b}{a} x^2+\frac{c}{a} x+\frac{d}{a}=0
$
Let the roots be $\alpha, \beta, \gamma$
$
\begin{aligned}
& \alpha+\beta+\gamma=-\frac{b}{a} \\
& \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a} \\
& \alpha \beta \gamma=-\frac{d}{a}
\end{aligned}
$
To find:
$
\begin{aligned}
\sum \frac{\alpha}{\beta \gamma} & =\frac{\alpha}{\beta \gamma}+\frac{\beta}{\gamma \alpha}+\frac{\gamma}{\alpha \beta}=\frac{\alpha^2+\beta^2+\gamma^2}{\alpha \beta \gamma} \\
& =\frac{(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)}{\alpha \beta \gamma} \\
& =\frac{\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)}{-\frac{d}{a}}=\frac{\left(b^2-2 a c\right)}{a^2} \times \frac{-a}{d}=\frac{2 a c-b^2}{a d}
\end{aligned}
$

Question 8.
If $\alpha, \beta, \gamma$ and $\delta$ are the roots of the polynomial equation $2 x^4+5 x^3-7 x^2+8=0$, find a quadratic equation with integer coefficients whose roots are $\alpha+\beta+\gamma+\delta$ and $\alpha \beta \gamma \delta$.
Solution:
The given equation is $2 x^4+5 x^3-7 x^2+8=0$.
$
\div 2 \Rightarrow x^4+\frac{5}{2} x^3-\frac{7}{2} x^2+4=0
$
Let the roots be $\alpha, \beta, \gamma, \delta$
$
\begin{aligned}
& \alpha+\beta+\gamma+\delta=-\frac{5}{2} \\
& \alpha \beta \gamma \delta=-4
\end{aligned}
$
To form the quadratic equation with the given roots $\alpha+\beta+\gamma+\delta$, $\alpha \beta \gamma \delta$.
$
\begin{aligned}
& \mathrm{x}^2-\mathrm{x} \text { (S.O.R) }+\text { P.O. } \mathrm{R}=0 \\
& x^2-x\left(\frac{-5}{2}-4\right)+\left(\frac{-5}{2}\right)(-4)=0 \\
& \Rightarrow x^2-x\left(\frac{-13}{2}\right)+10=0 \\
& 2 \mathrm{x}^2+13 \mathrm{x}+20=0
\end{aligned}
$
Question 9.
If $\mathrm{p}$ and $\mathrm{q}$ are the roots of the equation $1 \mathrm{x}^2+\mathrm{nx}+\mathrm{n}=0$, show that $\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$
Solution:

The given equation is $1 \mathrm{x}^2+\mathrm{nx}+\mathrm{n}=0$.
$
\begin{aligned}
& \mathrm{p}+\mathrm{q}=-\frac{n}{l}, \mathrm{pq}=\frac{n}{l} \\
& \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} q}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{\left(\frac{-n}{l}\right)}{\sqrt{\frac{n}{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \\
&=\frac{-n \sqrt{l}}{l \cdot \sqrt{n}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\sqrt{n}}{\sqrt{l}}+\frac{\sqrt{n}}{\sqrt{l}}=0
\end{aligned}
$

Question 10.
If the equations $x^2+p x+q=0$ and $x^2+p^{\prime} x+q$ ' $=0$ have a common root, show that it must be equal to $\frac{p q^{\prime}-p^{\prime} q}{q-q^{\prime}}$ or $\frac{q-q^{\prime}}{p^{\prime}-p}$
Solution:
If $\alpha$ is the common root, then.
$\alpha^2+p \alpha+q=0$
$\alpha^2+p^{\prime} \alpha+q^{\prime}=0$
Subtracting $\alpha\left(p-p^{\prime}\right)=q^{\prime}-\mathrm{q}$ $\alpha=\frac{q^{\prime}-q}{p-p^{\prime}}=\frac{q-q^{\prime}}{p^{\prime}-p} \quad \ldots \ldots .(3)$
Eliminating $\alpha$ from (1) \& (2)

Question 11.
Formulate into a mathematical problem to find a number such that when its cube root is added to it, the result is 6 .
Solution:
Let the number be $\mathrm{x}$.
Given that $\sqrt[3]{x}+x=6$
$
\Rightarrow \sqrt[3]{x}=6-x
$
Cubing on both sides
$
x=(6-x)^3
$
$
\begin{aligned}
& \Rightarrow \mathrm{x}=216-3(6)^2(\mathrm{x})+3(6)(\mathrm{x})^2-\mathrm{x}^3 \\
& \Rightarrow \mathrm{x}=216-108 \mathrm{x}+18 \mathrm{x}^2-\mathrm{x}^3 \\
& \Rightarrow \mathrm{x}^3-18 \mathrm{x}^2+109 \mathrm{x}-216=0
\end{aligned}
$
Question 12.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Solution:
Let the two parts be $x$ and $(12-x)$
Given that $x=\sqrt[3]{12-x}$
Cubing on both side,
$
\begin{aligned}
& \mathrm{x}^3=12-\mathrm{x} \\
& \Rightarrow \mathrm{x}^3+\mathrm{x}-12=0
\end{aligned}
$