Exercise 3.10 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 3.10$
Question $1 .$
Solve the following quadratic equations by factorization method
(i) $4 \mathrm{x}^{2}-7 \mathrm{x}-2=0$
(ii) $3\left(\mathrm{p}^{2}-6\right)=\mathrm{p}(\mathrm{p}+5)$
(iii) $\sqrt{a(a-7)}=3 \sqrt{2}$
(iv) $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$
(v) $2 \mathrm{x}^{2}-\mathrm{x}+\frac{1}{8}=0$
Solution:
(i)
$\begin{aligned}
4 x^{2}-7 x-2 &=0 \\
4 x^{2}-8 x+1 x-2 &=0 \\
4 x(x-2)+1(x-2) &=0 \\
\Rightarrow \quad(x-2)(4 x+1) &=0 \Rightarrow(x-2)=0 \\
x &=2 \text { or } 4 x+1=0 \\
x &=-\frac{1}{4}
\end{aligned}$
(ii) $3\left(\mathrm{p}^{2}-6\right)=\mathrm{p}(\mathrm{p}+5)$
$\begin{aligned}
&3 \mathrm{p}^{2}-18=\mathrm{p}^{2}+5 \mathrm{p} \Rightarrow 39^{2}-5 \mathrm{p}-18=0 \\
&\Rightarrow 2 \mathrm{p}^{2}-5 \mathrm{p}-18=0 \\
&\Rightarrow(2 \mathrm{p}-9)(\mathrm{p}+2)=0 \Rightarrow \mathrm{p}=\frac{9}{2},-2
\end{aligned}$
(iii) $\sqrt{a(a-7)}=3 \sqrt{2}$
Squaring on both sides
$a(a-7)=9 \times 2$

$\begin{aligned}
&a^{2}-7 a-18=0 \\
&a^{2}-9 a+2 a-18=0 \\
&a(a-9)+2(a-9)=0 \\
&(a-9)(a+2)=0 \\
&\Rightarrow a=9, a=-2
\end{aligned}$
(iv)
$\begin{aligned}
\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0 & \\
\sqrt{2} x^{2}+2 x+5 x+5 \sqrt{2} &=0 \\
\sqrt{2} x(x+\sqrt{2})+5(x+\sqrt{2}) &=0 \\
\Rightarrow \quad(x+\sqrt{2})(\sqrt{2} x+5) &=0 \\
x &=-\sqrt{2} \\
x &=\frac{-5}{\sqrt{2}}
\end{aligned}$
$\begin{aligned}
&\text { (v) } 2 x^{2}-x+\frac{1}{8}=0 \\
&16 x^{2}-8 x+1=0 \\
&16 x^{2}-4 x-4 x+1=0 \\
&4 x(4 x-1)-1(4 x-1)=0 \\
&(4 x-1)(4 x-1)=0 \\
&\Rightarrow x=\frac{1}{4}, \frac{1}{4}
\end{aligned}$

Question $2 .$
The number of volleyball games that must be scheduled in a league with $\mathrm{n}$ teams is given by $\mathrm{G}(\mathrm{n})$ $=\frac{n^{2}-n}{2}$ where each team plays with every other team exactly once. A league schedules 15 games.
How many teams are in the league?
Answer:
Number of games $=15$
$\begin{aligned}
&\mathrm{G}(\mathrm{n})=\frac{n^{2}-n}{2} \\
&\frac{n^{2}-n}{2}=15 \\
&\mathrm{n}^{2}-\mathrm{n}=30 \Rightarrow \mathrm{n}^{2}-\mathrm{n}-30=0
\end{aligned}$

$\begin{aligned}
&\Rightarrow n^{2}-6 n-5 n-30=0 \\
&(n-6)(n+5)=0 \\
&n-6=0 \text { or } n+5=0
\end{aligned}$
[Note: $-5$ is neglected because number of team is not negative]
$\mathrm{n}=6 \text { or } \mathrm{n}=-5$
$\therefore$ Number of teams $=6$