Exercise 3.5 - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.5$
Question 1.
Solve the following equations:
(i) $\sin ^2 x-5 \sin x+4=0$
(ii) $12 \mathrm{x}^3+8 \mathrm{x}=29 \mathrm{x}^2-4$
Solution:
(i) $\sin ^2 x-5 \sin x+4=0$
Let $y=\sin \mathrm{x}$
$\left(y^2-5 y+4=0\right.$
$(\mathrm{y}-1)(\mathrm{y}-4)=0$
$(y-1)=$ o or $(y-4)=0$
$y=1$ or $y=4$
$\sin \mathrm{x}=1$ or $\sin \mathrm{x}=4$ [not possible since $\sin \mathrm{x} \leq 1$ ]
$\sin \mathrm{x}=\sin \frac{\pi}{2}$
$\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \mathrm{n} \in \mathrm{z}$
$\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}$
(ii) $12 \mathrm{x}^3+8 \mathrm{x}=29 \mathrm{x}^2-4$
$12 \mathrm{x}^3-29 \mathrm{x}^2+8 \mathrm{x}+4=0$
By Trail and error method, $(x-2)$ is a factor of (1)
The other factor is $12 x^2-5 x-2$
The roots is $12 x^2-5 x-2=0$
$(3 \mathrm{x}-2)(4 \mathrm{x}+1)=0$
$\mathrm{x}=\frac{2}{3}, \mathrm{x}=-\frac{1}{4}$
The roots are $2, \frac{2}{3},-\frac{1}{4}$

[Here $\mathrm{a}_{\mathrm{n}}=12, \mathrm{a}_0=4 ;$ Let $\frac{p}{q}$ be the root of the equation (1) The factors of $\mathrm{a}_0: \pm 1, \pm 2, \pm 4$ (P must divisible by 4) The factor of $\mathrm{a}_{\mathrm{n}}: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ $\mathrm{q}$ must divide as (12)

Using these $\mathrm{p}$ and $\mathrm{q}$ we can form $\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{2}{3}, \pm \frac{1}{6}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm 3$ are the possible roots of equation. (1)]
Question 2.
Examine for the rational roots of:
(i) $2 x^3-x^2-1=0$
(ii) $x^8-3 x+1=0$
Solution:
(i) $2 x^3-x^2-1=0$
Sum of the co-efficients $=0$
$\therefore(\mathrm{x}-1)$ is a factor
The other factor is $2 x^2+x+1$.
The root is $\left(2 x^2+x+1\right)=0$
Here $\Delta=b^2-4 a c=(1)^2-4(2)(1)=1-8=-7<0$
The remaining roots are imaginary.
The only rational root is $\mathrm{x}=1$
(ii) $\mathrm{x}^8-3 \mathrm{x}+1=0$
Here $a_n=1, a_0=1$
If $\frac{p}{q}$ is a rational root of (1)
Then $\mathrm{q}$ is a factor $\mathrm{a}_{\mathrm{n}}, \mathrm{p}$ is a factor of $\mathrm{a}_0$
The possible values of $p$ and $q$ are $\pm 1$.
Among the possible values $1,-1,[(p, q)=1]$
None of them satisfies the equation (1)
The above equation has no rational roots.
Question 3.
Solve: $8 x^{\frac{3}{2 n}}-8 x^{\frac{-3}{2 n}}=63$
Solution:

(i)
$
\begin{aligned}
& 8 x^{\frac{3}{2 n}}-8 x^{\frac{-3}{2 n}}=63 \\
& \text { Let } y=x^{\frac{3}{2 n}} \\
& 8 y-\frac{8}{y}=63 \quad \Rightarrow \frac{8 y^2-8}{y}=63 \\
& 8 y^2-63 y-8=0 \\
& (8 y+1)(y-8)=0 \\
& (8 y+1)=0 \\
& (y-8)=0 \\
& y=\frac{-1}{8} \\
& y=8 \\
& x^{\frac{3}{2 n}}=\frac{-1}{8} \\
& x^{\frac{3}{2 n}}=8 \\
&
\end{aligned}
$
Squaring on both sides
$
\begin{array}{ll}
x=\left(\frac{1}{64}\right)^{\frac{n}{3}} & x=(64)^{\frac{n}{3}} \\
x=\left[\left(\frac{1}{4}\right)^3\right]^{\frac{n}{3}} & x=\left(4^3\right)^{\frac{n}{3}} \\
x=\frac{1}{4^n} \text { (not possible) } & x=4^n
\end{array}
$
only possible solution is $\mathrm{x}=4^{\mathrm{n}}$

Question 4.
Solve: $2 \sqrt{\frac{x}{a}}+3 \sqrt{\frac{a}{x}}=\frac{b}{a}+\frac{6 a}{b}$
Solution:

$\begin{gathered}
\text { Let } y=\sqrt{\frac{x}{a}} \\
\frac{2 y^2+3}{y}=\frac{6 a^2+b^2}{a b} \\
\left(2 y^2+3\right) a b=y\left(6 a^2+b^2\right) \\
2 a b y^2-y\left(6 a^2+b^2\right)+3 a b=0 \\
2 a b y^2-6 a^2 y-b^2 y+3 a b=0 \\
2 a y(b y-3 a)-b(b y-3 a)=0 \\
(2 a y-b)(b y-3 a)=0
\end{gathered}$

$
\begin{aligned}
(2 a y-b) & =0 \\
2 a y & =b \\
y & =\frac{b}{2 a} \\
\sqrt{\frac{x}{a}} & =\frac{b}{2 a}
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
\frac{x}{a} & =\frac{b^2}{4 a^2} \\
x & =\frac{b^2}{4 a}
\end{aligned}
$

$\begin{aligned}
(b y-3 a) & =0 \\
b y & =3 a \\
y & =\frac{3 a}{b} \\
\sqrt{\frac{x}{a}} & =\frac{3 a}{b} \\
\frac{x}{a} & =\frac{9 a^2}{b^2} \\
x & =\frac{9 a^3}{b^2}
\end{aligned}$

Question 5.
Solve the equations:
(i) $6 x^4-35 x^3+62 x^2-35 x+6=0$
(ii) $x^4+3 x^3-3 x-1=0$
Solution:
(i) $6 x^4-35 x^3+62 x^2-35 x+6=0$
It is a even degree reciprocal equation as $\mathrm{p}(\mathrm{x})=x^n p\left(\frac{1}{x}\right)$
Dividing equation (1) by $x^2$,

$\begin{array}{rlrl}
6 x^2-35 x+62-\frac{35}{x}+\frac{6}{x^2}=0 & & \Rightarrow 6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0 \\
\text { Let } y=x+\frac{1}{x} & & \Rightarrow x^2+\frac{1}{x^2}=\left(y^2-2\right) \\
6\left(y^2-2\right)-35 y+62=0 & \Rightarrow 6 y^2-12-35 y+62=0 \\
6 y^2-35 y+50=0 & \Rightarrow(3 y-10)(2 y-5)=0 \\
3 y-10=0,2 y-5=0 &
\end{array}$

$\text { Case (i): } \begin{aligned}
3\left(x+\frac{1}{x}\right) & =10 & & \Rightarrow 3\left(x^2+\frac{1}{x}\right)=10 \\
3 x^2+3 & =10 x & & \Rightarrow 3 x^2-10 x+3=0 \\
(3 x-1)(x-3) & =0 & & \Rightarrow x=\frac{1}{3}, x=3
\end{aligned}$

Case (ii):
$
\begin{array}{rlrl}
2 y-5 & =0 & & \Rightarrow 2 y=5 \\
2\left(x+\frac{1}{x}\right) & =5 & & \Rightarrow 2\left(x^2+1\right)=5 x \\
2 x^2-5 x+2 & =0 & & \Rightarrow(2 x-1)(x-2)=0 \\
x & =\frac{1}{2}, x=2 &
\end{array}
$
The roots are, $3, \frac{1}{3}, 2, \frac{1}{2}$

(ii) $x^4+3 x^3-3 x-1=0$
It is an even degree reciprocal function of type II.
$1,-1$ are the solution of equation (1)
$(\mathrm{x}-1),(\mathrm{x}+1)$ are the factor of (1)
$\left(x^2-1\right)$ is a factor of (1)
Dividing (1) by $\left(x^2-1\right)$
we get, $x^2+3 x+1=0$ is the other factor.

$
\begin{aligned}
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-3 \pm \sqrt{9-4}}{2} \\
& =\frac{-3 \pm \sqrt{5}}{2}
\end{aligned}
$
The roots are $1,-1, \frac{-3 \pm \sqrt{5}}{2}$

Question 6.
Find all real numbers satisfying $4^x-3\left(2^{x+2}\right)+2^5=0$
Solution:
$
\begin{aligned}
& 4^x-3\left(2^{x+2}\right)+2^5=0 \\
& \Rightarrow\left(2^2\right)^x-3\left(2^x \cdot 2^2\right)+2^5=0 \\
& \left(2^2\right)^x-12 \cdot 2^x+32=0
\end{aligned}
$
Let $y=2^x$
$
\begin{aligned}
& y^2-12 y+32=0 \\
& \Rightarrow(y-4)(y-8)=0 \\
& y-4=0 \text { or } y-8=0
\end{aligned}
$
Case (i): $2^{\mathrm{x}}=4$
$
\begin{aligned}
& \Rightarrow 2^{\mathrm{x}}=(2)^2 \\
& \Rightarrow \mathrm{x}=2
\end{aligned}
$
Case (ii): $2 x=8$
$
\begin{aligned}
& \Rightarrow 2^{\mathrm{x}}=(2)^3 \\
& \Rightarrow \mathrm{x}=3
\end{aligned}
$
$\therefore$ The roots are 2,3

Question 7.
Solve the equation $6 x^4-5 x^3-38 x^2-5 x+6=0$ if it is known that $\frac{1}{3}$ is a solution.
Solution:
$
\begin{aligned}
& 6 x^4-5 x^3-38 x^2-5 x+6=0 \\
& x=\frac{1}{3} \text { is a Solution } \\
& \therefore(3 x-1) \text { is a factor of }(1)
\end{aligned}
$
(1) is a Reciprocal equation even degree divide (1) by $x^2$.
$
\begin{aligned}
& 6 x^2-5 x-38-\frac{5}{x}+\frac{6}{x^2}=0 \\
& 6\left(x^2+\frac{1}{x}\right)-5\left(x+\frac{1}{x}\right)-38=0 \\
& \text { Let } y=x+\frac{1}{x}, x^2+\frac{1}{x^2}=y^2-2 \\
& 6\left(y^2-2\right)-5 y-38=0 \quad \Rightarrow 6 y^2-12-5 y-38=0 \\
& 6 y^2-5 y-50=0 \quad \Rightarrow(2 y+5)(3 y+10)=0 \\
& 2 y+5=0 ; 3 y-10=0 \\
&
\end{aligned}
$

Case (i)
$
\begin{aligned}
2 y & =-5 & & \\
2\left(x+\frac{1}{x}\right) & =-5 & & \Rightarrow 2\left(x^2+1\right)=-5 x \\
2 x^2+5 x+2 & =0 & & \Rightarrow(2 x+1)(x+2)=0 \\
x & =\frac{-1}{2}, x=-2 & &
\end{aligned}
$
Case (ii)
$
\begin{aligned}
3 y-10 & =0 & & \\
3\left(x+\frac{1}{x}\right) & =10 & & \Rightarrow 3 x^2+3=10 x \\
3 x^2-10 x+3 & =0 & & \Rightarrow(3 x-1)(x-3)=0 \\
x & =\frac{1}{3}, x=3 & &
\end{aligned}
$
$\therefore$ The roots are $3, \frac{1}{3}, 2, \frac{1}{2}$