Exercise 3.2 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.2$
Question $1 .$
Find the value of the polynomial $f(y)=6 y-3 y^{2}+3$ at (i) $y=1$ (ii) $y=-1$ (iii) $y=0$ Solution:
(i) At $y=1$,
$f(1)=6(1)-3(1)^{2}+3=6-3+3=6$
(ii) At $y=-1$,
$f(-1)=6(-1)-3(-1)^{2}+3=-6-3+3=-6$
(iii) At $\mathrm{y}=0$,
$f(0)=6(0)-3(0)^{2}+3=0-0+3=3$
 

Question 2.
If $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-2 \sqrt{2} x+1$, find $\mathrm{p}(2 \sqrt{2})$
Solution:
$\begin{aligned}
&\mathrm{p}(2 \sqrt{2})=(2 \sqrt{2})^{2}-2 \sqrt{2}(2 \sqrt{2})+1 \\
&=4 \times 2-4 \times 2+1 \\
&=8-8+1=1
\end{aligned}$

Question 3 .
Find the zeros of the polynomial in each of the following :
(i) $\mathrm{p}(\mathrm{x})=\mathrm{x}-3$
(ii) $\mathrm{p}(\mathrm{x})=2 \mathrm{x}+5$
(iii) $q(y)=2 y-3$,
(iv) $f(z)=8 z$
(v) $p(x)=$ ax where $a \neq 0$,
(vi) $h(x)=a x+b, a \neq 0$, a, b $\in R$
Solution:
(i) $x=3$.
$\mathrm{p}(3)=3-3=0$

$\therefore$ The zero of the polynomial is $\mathrm{x}=3$.
(ii) $p(x)=2 x+5, \quad$ if $2 x+5=0$.
$\begin{array}{rlrl}\therefore p\left(\frac{-5}{2}\right) & =22\left(\frac{-5}{2}\right)+5 & 2 x & =-5 \\ & =-5+5=0 & x & =\frac{-5}{2}\end{array}$
$\therefore x=\frac{-5}{2}$ is the zero of the polynomial $p(x)=2 x+5$.
(iii) $q(y)=2 y-3, \quad$ if $2 y-3=0$.
$\therefore \quad q\left(\frac{3}{2}\right)=\not{2}\left(\frac{3}{2}\right)-3=0 \quad \begin{aligned} 2 y &=3 \\ & \therefore \end{aligned}$
$\therefore y=\frac{3}{2}$ is the zero of the given polynomial.

(iv) $f(z)=8 z$,
If $8 z=0$
$\mathrm{z}=\frac{0}{8}=0$
$\mathrm{f}(0)=8(0)=0$
$\therefore \mathrm{z}=0$ is the zero of the given polynomial.
(v) $p(x)=a x$ when $a \neq 0$
$a x=0$
$p(0)=a(0)=0$
$x=\frac{0}{a}$
$\therefore x=0$ is the zero of the given polynomial. $\quad a$ $x=0$
(vi) $h(x)=a x+b, a \neq 0, a, b \in \mathbb{R}$
$\begin{array}{rlrl}
\therefore h\left(\frac{-b}{a}\right) & =a\left(\frac{-b}{a}\right)+b & \text { if } a x+b & =0 \\
& =-b+b=0 & a x & =-b \\
x & =\frac{-b}{a}
\end{array}$
$\therefore x=\frac{-b}{a}$ is the zero of the given polynomial.
 

Question $4 .$
Find the roots of the polynomial equations.
(i) $5 x-6=0$
(ii) $\mathrm{x}+3=0$
(iii) $10 \mathrm{x}+9=0$
(iv) $9 \mathrm{x}-4=0$

Solution:
(i) $5 x-6=0$
$\begin{aligned}
&5 x=6 \\
&\therefore x=\frac{6}{5}
\end{aligned}$
(ii) $x+3=0$
$\therefore \mathrm{x}=-3$
(iii) $10 x+9=0$
$10 \mathrm{x}=-9$ $\therefore \mathrm{x}=\frac{-9}{10}$ (iv) $9 \mathrm{x}-4=0$ $9 \mathrm{x}=4$ $\therefore \mathrm{x}=\frac{4}{9}$
 

Question $5 .$
Verify whether the following are zeros of the polynomial indicated against them, or not.
(i) $\mathrm{p}(\mathrm{x})=2 \mathrm{x}-1, \mathrm{x}=\frac{1}{2}$
(ii) $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-1, \mathrm{x}=1$,
(iii) $\mathrm{p}(\mathrm{x})=\mathrm{ax}+\mathrm{b}, \mathrm{x}=\frac{-b}{a}$
(iv) $\mathrm{p}(\mathrm{x})=(\mathrm{x}+3)(\mathrm{x}-4), \mathrm{x}=4, \mathrm{x}=-3$
Solution:
(i) $\mathrm{p}(\mathrm{x})=2 \mathrm{x}-1, \mathrm{x}=\frac{1}{2}$
$\mathrm{p}\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)-1=1-1=0$
$\therefore \mathrm{x}=\frac{1}{2}$ is the zero of the given polynomial.
(ii) $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-1, \mathrm{x}=1$
$\mathrm{p}(\mathrm{l})=1^{3}-1=1-1=0$
$\therefore \mathrm{x}=1$ is the zero of the given polynomial

(iii) $\mathrm{p}(\mathrm{x})=\mathrm{ax}+\mathrm{b}, \mathrm{x}=\frac{-b}{a}$
$\mathrm{p}\left(\frac{-b}{a}\right)=\mathrm{a}\left(\frac{-b}{a}\right)+\mathrm{b}$
$=-b+b=0$
$\therefore \mathrm{x}=\frac{-b}{a}$ is the zero of the given polynomial.
(iv) $\mathrm{p}(\mathrm{x})=(\mathrm{x}+3)(\mathrm{x}-4), \mathrm{x}=4, \mathrm{x}=-3$
$\begin{aligned}
&\mathrm{p}(-3)=(-3+3)(-3-4)=0(-7)=0 \\
&\mathrm{p}(4)=(4+3)(4-4)=7(0)=0
\end{aligned}$
$\therefore \mathrm{x}=-3, \mathrm{x}=4$ are the zeros of the given polynomial.
 

Question 6.
Find the number of zeros of the following polynomial represented by their graphs.

Solution:
(i) The curve cuts the $\mathrm{x}$-axis at two points. $\therefore$ The equation has 2 zeros.
(ii) Since the curve cuts the $x$-axis at 3 different points. The number of zeros of the given curve is three.
(iii) Since the curve doesn't cut the $x$ axis. The number of zeros of the given curve is zero.
(iv) The curve cut the $\mathrm{x}$-axis at one point. $\therefore$ The equation has one zero.
(v) The curve cut the $\mathrm{x}$ axis at one point. $\therefore$ The equation has one zero.