Additional Questions - Chapter 3 - Algebra - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Solve the following system of linear equations in three variables. $x+y+z=6 ; 2 x+3 y+4 z=20$; $3 x+2 y+5 z=22$
Solution:
$\begin{aligned}
&x+y+z=6 \ldots \ldots \ldots \ldots .(1) \\
&2 x+3 y+4 z=20 \ldots \ldots \ldots \ldots .(2) \\
&3 x+2 y+5 z=22 \ldots \ldots \ldots \ldots .(3)
\end{aligned}$

Sub. $z=3$ in $(5) \Rightarrow y-2(3)=-4$
$y=2$
Sub. $y=2, z=3$ in (1), we get
$\begin{aligned}
&x+2+3=6 \\
&x=1 \\
&x=1, y=2, z=3
\end{aligned}$
 

Question 2
Using quadratic formula solve the following equations.
(i) $\mathrm{p}^{2} \mathrm{x}^{2}+\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right) \mathrm{x}-\mathrm{q}^{2}=0$
(ii) $9 \mathrm{x}^{2}-9(\mathrm{a}+\mathrm{b}) \mathrm{x}+\left(2 \mathrm{a}^{2}+5 \mathrm{ab}+2 \mathrm{~b}^{2}\right)=0$
Solution:
(i) $\mathrm{p}^{2} \mathrm{x}^{2}+\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right) \mathrm{x}-\mathrm{q}^{2}=0$
Comparing this with $a x^{2}+b x+c=0$, we have
$\begin{aligned}
&a=p^{2} \\
&b=p^{2}-q^{2}
\end{aligned}$

$\begin{aligned}
&\mathrm{c}=-\mathrm{q}^{2} \\
&\Delta=\mathrm{b}^{2}-4 \mathrm{ac} \\
&=\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right)-4 \times \mathrm{p}^{2} \times-\mathrm{q}^{2} \\
&=\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right)^{2}+4 \mathrm{p}^{2} \mathrm{q}^{2} \\
&=\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)^{2}>0
\end{aligned}$
So, the given equation has real roots given by $\alpha=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)+\left(p^{2}+q^{2}\right)}{2 p^{2}}=\frac{q^{2}}{p^{2}}$
$\begin{aligned}
&\beta=\frac{-b-\sqrt{\Delta}}{2 a}=\frac{-\left(p^{2}-q^{2}\right)-\left(p^{2}+q^{2}\right)}{2 p^{2}} \\
&=-1
\end{aligned}$
(ii) $9 \mathrm{x}^{2}-9(\mathrm{a}+\mathrm{b}) \mathrm{x}+\left(2 \mathrm{a}^{2}+5 \mathrm{ab}+2 \mathrm{~b}^{2}\right)=0$
Comparing this with $a x^{2}+b x+c=0$.
$\begin{aligned}
&\mathrm{a}=9 \\
&\mathrm{~b}=-9(\mathrm{a}+\mathrm{b}) \\
&\mathrm{c}=\left(2 \mathrm{a}^{2}+5 \mathrm{ab}+2 \mathrm{~b}^{2}\right) \\
&\Delta=\mathrm{B}^{2}-4 \mathrm{AC} \\
&\Rightarrow 81(\mathrm{a}+\mathrm{b})^{2}-36\left(2 \mathrm{a}^{2}+5 \mathrm{ab}+2 \mathrm{~b}^{2}\right) \\
&\Rightarrow 9 \mathrm{a}^{2}+9 \mathrm{~b}^{2}-18 \mathrm{ab} \\
&\Rightarrow 9(\mathrm{a}-\mathrm{b})^{2}>0
\end{aligned}$

$\therefore$ the roots are real and given by
$\begin{aligned}
\alpha &=\frac{-\mathrm{B}-\sqrt{\Delta}}{2 \mathrm{~A}}=\frac{9(a+b)+3(a-b)}{18} \\
&=\frac{12 a+6 b}{18}=\frac{2 a+b}{3} \\
\beta &=\frac{-\mathrm{B}-\sqrt{\Delta}}{2 \mathrm{~A}}=\frac{9(a+b)-3(a-b)}{18} \\
&=\frac{6 a+12 b}{18}=\frac{a+2 b}{3}
\end{aligned}$

Question $3 .$
Find the HCF of $x^{3}+x^{2}+x+1$ and $x^{4}-1$
Answer:
$\begin{aligned}
&x^{3}+x^{2}+x+1=x^{2}(x+1)+1(x+1) \\
&=(x+1)\left(x^{2}+1\right) \\
&x^{4}-1=\left(x^{2}\right)^{2}-1 \\
&=\left(x^{2}+1\right)\left(x^{2}-1\right) \\
&=\left(x^{2}+1\right)(x+1)(x-1) \\
&\text { H.C.F. }=\left(x^{2}+1\right)(x+1)
\end{aligned}$
 

Question 4
Prove that the equation $x^{2}\left(a^{2}+b^{2}\right)+2 x(a c+b d)+\left(c^{2}+d^{2}\right)=0$ has no real root if $a d \neq b c$
Solution:
$\begin{aligned}
&\Delta=b^{2}-4 a c \\
&\Rightarrow 4(a c+b d)^{2}-4\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
&\Rightarrow 4\left[(a c+b d)^{2}-\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\right] \\
&\Rightarrow 4\left(a^{2} c^{2}+b^{2} d^{2}+2 a c b d-a^{2} c^{2} b^{2} c^{2}-a^{2} d^{2}-b^{2} d^{2}\right] \\
&\Rightarrow 4\left[2 a c b d-a^{2} d^{2}-b^{2} c^{2}\right]
\end{aligned}$

$\begin{aligned}
&\Rightarrow 4\left[\mathrm{a}^{2} \mathrm{c}^{2}+\mathrm{b}^{2} \mathrm{c}^{2}-2 \mathrm{adbc}\right] \\
&\Rightarrow-4[\mathrm{ad}-\mathrm{bc}]^{2} \\
&\text { We have ad } \neq \mathrm{hc} \\
&\therefore \mathrm{ad}-\mathrm{bc}>0 \\
&\Rightarrow(\mathrm{ad}-\mathrm{bc})^{2}>0
\end{aligned}$
$\Rightarrow-4(\mathrm{ad}-\mathrm{bc})^{2}<0 \Rightarrow \Delta<0$
Hence the given equation has no real roots.
 

Question $5 .$
Find the L.C.M of $2\left(x^{3}+x^{2}-x-1\right)$ and $3\left(x^{3}+3 x^{2}-x-3\right)$
Answer:
$\begin{aligned}
&2\left[x^{3}+x^{2}-x-1\right]=2\left[x^{2}(x+1)-1(x+1)\right] \\
&=2(x+1)\left(x^{2}-1\right) \\
&=2(x+1)(x+1)(x-1) \\
&=2(x+1)^{2}(x-1) \\
&3\left[x^{3}+3 x^{2}-x-3\right]=3\left[x^{2}(x+3)-1(x+3)\right] \\
&=3\left[(x+3)\left(x^{2}-1\right)\right] \\
&=3(x+3)(x+1)(x-1) \\
&\text { L.C.M. }=6(x+1)^{2}(x-1)(x+3)
\end{aligned}$

Question 6.
A two digit number is such that the product of its digits is 12 . When 36 is added to the number the digits interchange their places. Find the number.
Solution:
Let the ten's digit of the number be $\mathrm{x}$. It is given that the product of the digits is 12 .

Unit's digit $=\frac{12}{x}$
Number $=10 \mathrm{x}+\frac{12}{x}$
If 36 is added to the number the digits interchange their places.
$\begin{aligned}
&\therefore 10 x+\frac{12}{x}+36=10 \times \frac{12}{x}+x \\
&\Rightarrow 10 x+\frac{12}{x}+36=\frac{120}{x}+x \\
&\Rightarrow 9 x-\frac{108}{x}+36=0 \\
&\Rightarrow 9 x^{2}-108+36 x=0 \\
&\Rightarrow \quad x^{2}+4 x-12=0 \\
&\Rightarrow(x+6)(x-2)=0 \quad(\because(x+6) \neq 0 \text { as } x>0) \\
&x=-6,2
\end{aligned}$
But a number can never be (-ve). So, $x=2$.
The number is $10 \times 2+\frac{12}{2}=26$
 

Question $7 .$
Seven years ago, Vanin's age was five times the square of swati's age. Three years hence Swati's age will be two fifth of Varun's age. Find their present ages.
Solution:
Seven years ago, let Swathi's age be $x$ years.
Seven years ago, let Varun's age was $5 x^{2}$ years.
Swathi's present age $=x+7$ years

Varun's present age $=\left(5 x^{2}+7\right)$ years
3 years hence, we have Swathi's age $=x+7+3$ years
$=\mathrm{x}+10$ years
Varun's age $=5 x^{2}+7+3$ years
$=5 \mathrm{x}^{2}+10$ years
It is given that 3 years hence Swathi's age will be $\frac{2}{5}$ of Varun's age.
$\begin{aligned}
&\therefore x+10=\frac{2}{5}\left(5 x^{2}+10\right) \\
&\Rightarrow x+10=2 x^{2}+4 \\
&\Rightarrow 2 x^{2}-x-6=0 \\
&\Rightarrow 2 x(x-2)+3(x-2)=0 \\
&\Rightarrow(2 x+3)(x-2)=0 \\
&\Rightarrow x-2=0 \\
&\Rightarrow x=2(\because 2 x+3 \neq 0 \text { as } x>0)
\end{aligned}$
Hence Swathi's present age $=(2+7)$ years
$=9$ years
Varun's present age $=\left(5 \times 2^{2}+7\right)$ years $=27$ years
 

Question 8 .
A chess board contains 64 equal squares and the area of each square is $6.25 \mathrm{~cm}^{2}$. A border round the board is $2 \mathrm{~cm}$ wide find its side.
Solution:
Let the length of the side of the chess board be $x \mathrm{~cm}$. Then,

Area of 64 squares $=(x-4)^{2}$
$\begin{aligned}
&(x-4)^{2}=64 \times 6.25 \\
&\Rightarrow x^{2}-8 x+16=400 \\
&\Rightarrow x^{2}-8 x-384=0 \\
&\Rightarrow x^{2}-24 x+16 x-384=0 \\
&\Rightarrow(x-24)(x+16)=0 \\
&\Rightarrow x=24 \mathrm{~cm}
\end{aligned}$

Question $9 .$
Find two consecutive natural numbers whose product is 20 .
Solution:
Let a natural number be $x$.
The next number $=x+1$
$\begin{aligned}
&x(x+1)=20 \\
&x^{2}+x-20=0 \\
&(x+5)(x-4)=0 \\
&x=-5,4 \\
&\therefore x=4(\because x \neq-5, x \text { is natural number })
\end{aligned}$
The next number $=4+1=5$
Two consecutive numbers are 4,5
 

Question $10 .$
A two digit number is such that the product of its digits is 18 , when 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution:

Let the tens digit be $x$. Then the units digits $=\frac{18}{x}$
$\therefore$ Number $=10 x+\frac{18}{x}$
and number obtained by interchanging the digits $=10 \times \frac{18}{x}+x$
$\therefore\left(10 x+\frac{18}{x}\right)-\left(10 \times \frac{18}{x}+x\right)=63$
$\Rightarrow \quad 10 x+\frac{18}{x}-\frac{180}{x}-x=0$
$\Rightarrow \quad 9 x-\frac{162}{x}-63=0$
$\Rightarrow \quad 9 x^{2}-63 x-162=0$
$\Rightarrow \quad x^{2}-7 x-18=0$
$\Rightarrow(x-9)(x+2)=0 \Rightarrow x=9,-2$
But a digit can never be (-ve), so $x=9$.
So, the required number $=10 \times 9+\frac{18}{9}=92$.