Exercise 3.14 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 3.14
Question 1.
Write each of the following expression in terms of $\alpha+\beta$ and $\alpha \beta$.
(i) $\frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha}$
(ii) $\frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha}$
(iii) $(3 \alpha-1)(3 \beta-1)$ (iv) $\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha}$
(i)
$\begin{aligned}
\frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha} &=\frac{\alpha^{2}+\beta^{2}}{3 \alpha \beta} \\
&=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{3 \alpha \beta}
\end{aligned}$
(ii)
$\begin{aligned}
\frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha} &=\frac{\beta+\alpha}{\alpha^{2} \beta^{2}} \\
&=\frac{\alpha+\beta}{(\alpha \beta)^{2}}
\end{aligned}$
(iii) $(3 \alpha-1)(3 \beta-1)=9 \alpha \beta-3 \beta-3 \alpha+1$
$=9 \alpha \beta-3(\alpha+\beta)+1$
(iv)
$\begin{aligned}
\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha} &=\frac{\alpha^{2}+3 \alpha+\beta^{2}+3 \beta}{\alpha \beta} \\
&=\frac{\alpha^{2}+\beta^{2}+3(\alpha+\beta)}{\alpha \beta}
\end{aligned}$

$=\frac{\alpha^{2}+\beta^{2}+3(\alpha+\beta)}{\alpha \beta}$
$=\frac{(\alpha+\beta)^{2}-2 \alpha \beta+3(\alpha+\beta)}{\alpha \beta}$

Question 2.
The roots of the equation $2 x^{2}-7 x+5=0$ are $\alpha$ and $\beta$. Without solving the root find
(i) $\frac{1}{\alpha}+\frac{1}{\beta}$
(ii) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
(iii) $\frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2}$
Solution:
$\begin{aligned}
&2 \mathrm{x}^{2}-7 \mathrm{x}+5=\mathrm{x}^{2}-\frac{7}{2} x+\frac{5}{2}=0 \\
&\alpha+\beta=\frac{7}{2} \\
&\alpha \beta=\frac{5}{2}
\end{aligned}$
(i) $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{\frac{7}{2}}{\frac{5}{2}}=\frac{7}{5}$
(ii)
$\begin{aligned}
\frac{\alpha}{\beta}+\frac{\beta}{\alpha} &=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta} \\
&=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta} \\
&=\frac{(\alpha+\beta)^{2}}{\alpha \beta}-2 \\
&=\frac{\left(\frac{7}{2}\right)^{2}}{\frac{5}{2}}-2=\frac{49}{4} \times \frac{2}{5}-2 \\
=\frac{49}{10}-2 &=\frac{49-20}{10}=\frac{29}{10}
\end{aligned}$

(iii) $\frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2}$
$\begin{aligned}
&=\frac{(\alpha+2)^{2}+(\beta+2)^{2}}{\alpha \beta+2 \alpha+2 \beta+4} \\
&=\frac{\alpha^{2}+4 \alpha+4+\beta^{2}+4 \beta+4}{\alpha \beta+2(\alpha+\beta)+4} \\
&=\frac{(\alpha+\beta)^{2}-2 \alpha \beta+4(\alpha+\beta)+8}{\alpha \beta+2(\alpha+\beta)+4} \\
&=\frac{\frac{49}{4}-\frac{10}{2}+\frac{28}{2}+\frac{16}{2}}{\frac{5}{2}+\frac{14}{2}+\frac{8}{2}} \\
&=\frac{49-20+56+32}{5+14+8} \times \frac{1}{2} \\
&=\frac{117}{54}
\end{aligned}$

Question $3 .$
The roots of the equation $x^{2}+6 x-4=0$ are $\alpha$, $\beta$. Find the quadratic equation whose roots are
(i) $\alpha^{2}$ and $\beta^{2}$
(ii) $\frac{2}{\alpha}$ and $\frac{2}{\beta}$
(iii) $\alpha^{2} \beta$ and $\beta^{2} \alpha$
Solution:
If the roots are given, the quadratic equation is $x^{2}-$ (sum of the roots) $x+$ product the roots $=0$.
For the given equation.

$\begin{aligned}
&\text { (i) } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta \\
&=(-6)^{2}-2(-4)=36+8=44 \\
&\alpha^{2} \beta^{2}=(\alpha \beta)^{2}=(-4)^{2}=16
\end{aligned}$
$\therefore$ The required equation is $x^{2}-44 x-16=0$.
(ii)
$\begin{gathered}
\frac{2}{\alpha}+\frac{2}{\beta}=\frac{2 \beta+2 \alpha}{\alpha \beta}=\frac{2(\alpha+\beta)}{\alpha \beta}=\frac{2(-6)}{-4} \\
=\frac{-12}{-4}=3 \\
\frac{2}{\alpha}+\frac{2}{\beta}=\frac{4}{\alpha \beta}=\frac{4}{-4}=-1
\end{gathered}$
$\therefore$ The required equation is $x^{2}-3 x-1=0$.
(iii) $\alpha^{2} \beta+\beta^{2} \alpha=\alpha \beta(\alpha+\beta)$
$=-4(-6)=24$
$\alpha^{2} \beta \times \beta^{2} \alpha=\alpha^{3} \beta^{3}=(\alpha \beta)^{3}=(-4)^{3}=-64$
$\therefore$ The required equation $=x^{2}-24 x-64-0$.
 

Question $4 .$
If $\alpha, \beta$ are the roots of $7 x^{2}+a x+2=0$ and if $\beta-\alpha=\frac{-13}{7}$ Find the values of $a$.

Solution:
$7 x^{2}+a x+2=0$
$x=\frac{-a \pm \sqrt{a^{2}-56}}{2 \times 7}$
$\alpha=\frac{-a+\sqrt{a^{2}-56}}{14}, \beta=\frac{-a-\sqrt{a^{2}-56}}{14}$
$\beta-\alpha=\frac{-a-\sqrt{a^{2}-56}+a-\sqrt{a^{2}-56}}{14}$
$=\frac{-2 \sqrt{a^{2}-56}}{14}=\frac{-13}{7}$ (given)
$\Rightarrow \frac{-\sqrt{a^{2}-56}}{7}=\frac{-13}{7}$
$\Rightarrow-\sqrt{a^{2}-56}=-13$
Squaring on both sides.
$\begin{aligned}
a^{2}-56 &=169 \\
a^{2} &=225 \\
a &=\pm 15
\end{aligned}$
 

Question $5 .$
If one root of the equation $2 y^{2}-a y+64=0$ is twice the other then find the values of a.
Solution:
Let one of the root $\alpha=2 \beta$
$\alpha+\beta=2 \beta+\beta=3 \beta$
Given

$\begin{aligned}
2 y^{2}-a y+64 &=0 \\
y^{2}-\frac{a}{2} y+32 &=0 \\
\Rightarrow y^{2}-\left(\frac{a}{2}\right) y+32 &=0
\end{aligned}$
Sum of the roots $\alpha+\beta=\frac{a}{2}$
$3 \beta=\frac{a}{2} \Rightarrow \beta=\frac{a}{6}$
$\alpha \beta=\alpha \times \frac{a}{6}$
$\Rightarrow, \quad 2 \beta \times \beta=2\left(\frac{a}{6}\right)\left(\frac{a}{6}\right)$
$(2 \beta \beta)=2 \beta^{2}=32$
$2\left(\frac{a^{2}}{36}\right)=36$
$a^{2}=576$
$a=24,=24$

Question $6 .$
If one root of the equation $3 x^{2}+k x+81=0$ (having real roots) is the square of the other then find $\mathrm{k}$
Solution:
$3 x^{2}+k x+81=0$

$\begin{aligned}
&\text { Let the roots be } \alpha \text { and } \alpha^{2}\\
&\alpha+\alpha^{2}=\frac{-k}{3}\\
&\alpha \alpha^{2}=\frac{81}{3}\\
&\Rightarrow \quad \alpha^{3}=27\\
&\Rightarrow \quad \alpha=3\\
&3+3^{2}=\frac{-k}{3}\\
&\Rightarrow \quad(3+9)=\frac{-k}{3}\\
&\Rightarrow \quad k=-36 \text {. }
\end{aligned}$