Koshe Dekhi 8.5 Class 9

উৎপাদকে বিশ্লেষণ

1. নিচের বহুপদী সংখ্যামালাগুলিকে করি :

(i) $\fn_cm {\color{Blue} \left ( a+b \right )^{2}-5a-5b+6}$

সমাধানঃ

$\fn_cm \left ( a+b \right )^{2}-5a-5b+6$

$\fn_cm =\left ( a+b \right )^{2}-5\left (a+b \right )+6$

$\fn_cm =\left ( a+b \right )^{2}-3\left (a+b \right )-2\left (a+b \right )+6$

$=\left ( a+b \right )\left [ \left ( a+b \right )-3 \right ]-2\left [ \left ( a+b \right )-3 \right ]$

$=\left ( a+b-3 \right )\left ( a+b-2 \right )$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(ii) $\fn_cm {\color{Blue} \left ( x+1 \right )\left ( x+2 \right )\left ( 3x-1 \right )\left ( 3x-4 \right )+12}$

সমাধানঃ

$\fn_cm \left ( x+1 \right )\left ( x+2 \right )\left ( 3x-1 \right )\left ( 3x-4 \right )+12$

$=\left [ \left ( x+1 \right )\left ( 3x-1 \right ) \right ]\left [ \left ( x+2 \right )\left ( 3x-4 \right ) \right ]+12$

$=\left [ 3x^{2}-x+3x-1 \right ]\left [ 3x^{2}-4x+6x-8 \right ]+12$

$=\left [ 3x^{2}+2x-1 \right ]\left [ 3x^{2}+2x-8 \right ]+12$

$=\left ( a-1 \right )\left ( a-8 \right )+12$ [ ধরি, ${\color{Blue} 3x^{2}+2x=a]}$

$=a^{2}-8a-a+8+12$

$=a^{2}-9a+20$

$=a^{2}-5a-4a+20$

$=a\left ( a-5 \right )-4\left ( a-5 \right )$

$=\left ( a-5 \right )\left ( a-4 \right )$

$=\left ( 3x^{2}+2x-5 \right )\left ( 3x^{2}+2x-4 \right )$ [ a এর মান বসিয়ে পাই ]

$=\left ( 3x^{2}+5x-3x-5 \right )\left ( 3x^{2}+2x-4 \right )$

$=\left [x\left ( 3x+5 \right )-1\left ( 3x+5 \right ) \right ]\left ( 3x^{2}+2x-4 \right )$

$=\left ( 3x+5 \right )\left ( x-1 \right )\left ( 3x^{2}+2x-4 \right )$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(iii) $\fn_cm {\color{Blue} x\left ( x^{2}-1 \right )\left ( x+2 \right )-8}$

সমাধানঃ

$\fn_cm x\left ( x^{2}-1 \right )\left ( x+2 \right )-8$

$=x\left [\left ( x+1 \right )\left ( x-1 \right ) \right ]\left ( x+2 \right )-8$

[ সূত্রঃ ${\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )}$ ]

$=x\left ( x+1 \right )\left ( x-1 \right ) \left ( x+2 \right )-8$

$=\left [x\left ( x+1 \right ) \right ]\left [\left ( x-1 \right ) \left ( x+2 \right ) \right ]-8$

$=\left ( x^{2}+x \right )\left ( x^{2}+2x-x-2 \right )-8$

$=\left ( x^{2}+x \right )\left ( x^{2}+x-2 \right )-8$

$=a\left ( a-2 \right )-8$ [ ধরি, ${\color{Blue} x^{2}+x=a}$ ]

$=a^{2}-2a-8$

$=a^{2}-4a+2a-8$

$=a\left ( a-4 \right )+2\left ( a-4 \right )$

$=\left ( a+2 \right )\left ( a-4 \right )$

$=\left ( x^{2}+x+2 \right )\left ( x^{2}+x-4 \right )$ [ a এর মান বসিয়ে পাই ]

উৎপাদকে বিশ্লেষণ

(iv) $\fn_cm {\color{Blue} 7\left ( a^{2}+b^{2} \right )^{2}-15\left ( a^{4}-b^{4} \right )+8\left ( a^{2}-b^{2} \right )^{2}}$

সমাধানঃ

$\fn_cm 7\left ( a^{2}+b^{2} \right )^{2}-15\left ( a^{4}-b^{4} \right )+8\left ( a^{2}-b^{2} \right )^{2}$

$=7\left ( a^{2}+b^{2} \right )^{2}-15\left [\left ( a^{2}+b^{2} \right )\left ( a^{2}-b^{2} \right ) \right ]+8\left ( a^{2}-b^{2} \right )^{2}$

[ সূত্রঃ ${\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )}$ ]

$=7p^{2}-15pq+8q^{2}$

[ ধরি, ${\color{Blue} \left ( a^{2}+b^{2} \right )=p,\left ( a^{2}-b^{2} \right )=q}$ ]

$=7p^{2}-7pq-8pq+8q^{2}$

$=7p\left ( p-q \right )-8q\left ( p-q \right )$

$=\left ( p-q \right )\left ( 7p-8q \right )$

$=\left ( a^{2}+b^{2}-a^{2}+b^{2} \right )\left [ 7\left ( a^{2}+b^{2} \right )-8\left (a^{2}-b^{2} \right ) \right ]$ [ p ও q এর মান বসিয়ে পাই ]

$=2b^{2}\left ( 7a^{2}+7b^{2}-8a^{2}+8b^{2} \right )$

$=2b^{2}\left ( 15b^{2}-a^{2} \right )$

উৎপাদকে বিশ্লেষণ

(v) $\fn_cm {\color{Blue} \left ( x^{2}-1 \right )^{2}+8x\left ( x^{2}+1 \right )+19x^{2}}$

সমাধানঃ

$\fn_cm \left ( x^{2}-1 \right )^{2}+8x\left ( x^{2}+1 \right )+19x^{2}$

$\fn_cm =\left ( x^{2}+1 \right )^{2}-4x^{2}+8x\left ( x^{2}+1 \right )+19x^{2}$

${\color{Blue} \left [ \because \left ( a-b \right )^{2}=\left ( a+b \right )^{2}-4ab \right ]}$

$\fn_cm =\left ( x^{2}+1 \right )^{2}+8x\left ( x^{2}+1 \right )+15x^{2}$

$\fn_cm =\left ( x^{2}+1 \right )^{2}+5x\left ( x^{2}+1 \right )+3x\left ( x^{2}+1 \right )+15x^{2}$

$\fn_cm =\left ( x^{2}+1 \right )\left [\left ( x^{2}+1 \right )+5x \right ]+3x\left [\left ( x^{2}+1 \right )+5x \right ]$

$=\left ( x^{2}+1+3x \right )\left ( x^{2}+1+5x \right )$

$\fn_cm =\left ( x^{2}+3x+1 \right )\left ( x^{2}+5x+1 \right )$

উৎপাদকে বিশ্লেষণ

(vi) $\fn_cm {\color{Blue} \left ( a-1 \right )x^{2}-x-\left ( a-2 \right )}$

সমাধানঃ

$\fn_cm \left ( a-1 \right )x^{2}-x-\left ( a-2 \right )$

$=ax^{2}-x^{2}-x-a+2$

$=ax^{2}-x^{2}-x-a+1+1$

$=ax^{2}-a-x^{2}+1-x+1$

$=a\left ( x^{2}-1 \right )-\left ( x^{2}-1 \right )-1\left ( x-1 \right )$

$=a\left ( x+1 \right )\left ( x-1 \right )-\left ( x+1 \right )\left ( x-1 \right )-1\left ( x-1 \right )$

[ সূত্রঃ ${\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )}$ ]

$=\left ( x-1 \right )\left [ a\left ( x+1 \right )-\left ( x+1 \right ) -1\right ]$

$=\left ( x-1 \right )\left ( ax+a-x-1-1 \right )$

$=\left ( x-1 \right )\left ( ax+a-x-2 \right )$

উৎপাদকে বিশ্লেষণ

(vii) $\fn_cm {\color{Blue} \left ( a-1 \right )x^{2}+a^{2}xy+\left ( a+1 \right )y^{2}}$

সমাধানঃ

$\left ( a-1 \right )x^{2}+a^{2}xy+\left ( a+1 \right )y^{2}$

$=\left ( a-1 \right )x^{2}+\left [ \left ( a^{2}-1 \right )+1 \right ]xy+\left ( a+1 \right )y^{2}$

$=\left ( a-1 \right )x^{2}+ \left ( a^{2}-1 \right )xy+xy+\left ( a+1 \right )y^{2}$

$=\left ( a-1 \right )x\left [ x+\left ( a+1 \right )y \right ]+y\left [ x+\left ( a+1 \right )y \right ]$

$=\left ( ax-x \right )\left [ x+ay+y \right ]+y\left [ x+ay+y \right ]$

$=\left ( x+ay+y \right )\left ( ax-x+y \right )$

Koshe dekhi 8.5 class 9

উৎপাদকে বিশ্লেষণ

(viii) $\fn_cm {\color{Blue} x^{2}-qx-p^{2}+5pq-6q^{2}}$

সমাধানঃ

$\fn_cm x^{2}-qx-p^{2}+5pq-6q^{2}$

$\fn_cm =x^{2}-qx-\left (p^{2}-5pq+6q^{2} \right )$

$\fn_cm =x^{2}-qx-\left (p^{2}-3pq-2pq+6q^{2} \right )$

$\fn_cm =x^{2}-qx-\left [p\left (p-3q \right )-2q\left ( p-3q \right ) \right ]$

$\fn_cm =x^{2}-qx-\left (p-3q \right )\left ( p-2q \right )$

$\fn_cm =x^{2}-\left [ \left ( p-2q \right )-\left ( p-3q \right ) \right ]x-\left (p-3q \right )\left ( p-2q \right )$

$\fn_cm =x^{2}-\left ( p-2q \right )x+\left ( p-3q \right )x-\left (p-3q \right )\left ( p-2q \right )$

$=x\left ( x-p+2q \right )+\left ( p-3q \right )\left ( x-p+2q \right )$

$=\left ( x+p-3q \right )\left ( x-p+2q \right )$

উৎপাদকে বিশ্লেষণ

(ix) $\fn_cm {\color{Blue} 2\left ( a^{2}+\frac{1}{a^{2}} \right )-\left ( a-\frac{1}{a} \right )-7}$

সমাধানঃ

$\fn_cm {\color{Blue} 2\left ( a^{2}+\frac{1}{a^{2}} \right )-\left ( a-\frac{1}{a} \right )-7}$

$\fn_cm =2\left [\left ( a-\frac{1}{a} \right )^{2}+2 \right ]-\left ( a-\frac{1}{a} \right )-7$

ধরি, ${\color{Blue} \left ( a-\frac{1}{a} \right )=x}$

$\fn_cm =2\left ( x^{2}+2 \right ) -x-7$

$\fn_cm =2x^{2}+4-x-7$

$\fn_cm =2x^{2}-x-3$

$\fn_cm =2x^{2}-3x+2x-3$

$=x\left ( 2x-3 \right )+1\left ( 2x-3 \right )$

$=\left ( 2x-3 \right )\left ( x+1 \right )$

x এর মান বসিয়ে পাই,

$=\left [ 2\left ( a-\frac{1}{a}-3 \right ) \right ]\left ( a-\frac{1}{a}+1 \right )$

$=\left ( 2a-\frac{2}{a}-3 \right )\left ( a-\frac{1}{a}+1 \right )$

উৎপাদকে বিশ্লেষণ

(x) $\fn_cm {\color{Blue} \left ( x^{2}-x \right )y^{2}+y-\left ( x^{2}+x \right )}$

সমাধানঃ

$\fn_cm \left ( x^{2}-x \right )y^{2}+y-\left ( x^{2}+x \right )$

$\fn_cm =x\left ( x-1 \right )y^{2}+y-x\left ( x+1 \right )$

$=x\left ( x-1 \right )y^{2}+\left [ x^{2}-\left \{ \left ( x+1 \right )\left ( x-1 \right ) \right \} \right ]y-x\left ( x+1 \right )$

$=x\left ( x-1 \right )y^{2}+x^{2}y-\left ( x+1 \right )\left ( x-1 \right ) y-x\left ( x+1 \right )$

$=xy\left [ \left ( x-1 \right )y+x \right ]-\left ( x+1 \right )\left [ \left ( x-1 \right )y+x \right ]$

$=\left [ \left ( x-1 \right )y+x \right ]\left [xy-\left ( x+1 \right ) \right ]$

$=\left ( xy-y+x \right )\left ( xy-x-1 \right )$

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(i) $\fn_cm {\color{Blue} a^{2}-b^{2}=11\times 9}$ এবং a ও b ধনাত্মক পূর্ণসংখ্যা (a > b) হলে

(a) a = 11, b = 9

(b) a = 33, b = 3

(d) a = 100, b = 1

সমাধানঃ

$\fn_cm a^{2}-b^{2}=11\times 9$

বা, $\left ( a+b \right )\left ( a-b \right )=\left ( 10+1 \right )\left ( 10-1 \right )$

উভয়পক্ষ তুলনা করে পাই,

$a=10,b=1$

উত্তরঃ (c) a = 10, b = 1

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(ii) যদি $\fn_cm {\color{Blue} \frac{a}{b}+ \frac{b}{a}=1}$ হয়, তাহলে $\fn_cm {\color{Blue} a^{3}+b^{3}}$ -এর মান

(a) 1

(b) a

(d) 0

সমাধানঃ

$\fn_cm \frac{a}{b}+ \frac{b}{a}=1$

বা, $\frac{a^{2}+b^{2}}{ab}=1$

বা, $a^{2}+b^{2}=ab$

বা, $a^{2}+b^{2}-ab=0$

বা, $\left ( a+b \right )\left (a^{2}+b^{2}-ab \right )=0\times \left ( a+b \right )$

বা, $a^{3}+b^{3}=0$

উত্তরঃ (d) 0

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iii) $\fn_cm {\color{Blue} 25^{3}-75^{3}+50^{3}+3\times 25\times 75\times 50}$ -এর মান

(a) 150

(b) 0

(d) 50

সমাধানঃ

$\fn_cm 25^{3}-75^{3}+50^{3}+3\times 25\times 75\times 50$

$=\left ( 25-75+50 \right )\left [ \left ( 25 \right )^{2}+\left ( -75 \right )^{2}+\left ( 50 \right )^{2}-25\times \left ( -75 \right )-\left ( -75 \right )\times 50-50\times 25 \right ]$

${\color{Blue} \left [ \because a^{3}+b^{3}+c^{3}+3abc=\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )\right ]}$

= 0

উত্তরঃ (b) 0

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(iv) $\fn_cm {\color{Blue} a+b+c=0}$ হলে $\fn_cm {\color{Blue} \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}}$ -এর মান

(a) 0

(b) 1

(d) 3

সমাধানঃ

$\fn_cm \frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}$

$=\frac{a^{3}+b^{3}+c^{3}}{abc}$

$=\frac{\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )+3abc}{abc}$

$=\frac{0\times \left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )}{abc}+\frac{3abc}{abc}$

= 0 + 3

= 3

উত্তরঃ (d) 3

উৎপাদকে বিশ্লেষণ

2. বহু বিকল্পীয় প্রশ্নঃ (M.C.Q.):

(v) $\fn_cm {\color{Blue} x^{2}-px+12=\left ( x-3 \right )\left ( x-a \right )}$ একটি অভেদ হলে, a ও p -এর মান যথাক্রমে

(a) a = 4, p = 7

(b) a = 7, p = 4

(d) a = −4, p = 7

সমাধানঃ

$\fn_cm x^{2}-px+12=\left ( x-3 \right )\left ( x-a \right )$

বা, $\fn_cm x^{2}-px+12=x^{2}-3x-ax+3a$

বা, $\fn_cm x^{2}-px+12=x^{2}-x\left (3+a \right )+3a$

উভয়পক্ষ তুলনা করে পাই,

$p=3+a,3a=12$

$\therefore a=\frac{12}{3}=4$

ও $p=3+4=7$

উত্তরঃ (a) a = 4, p = 7

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(i) $\fn_cm {\color{Blue} \frac{\left ( b^{2}-c^{2} \right )^{3}+\left ( c^{2}-a^{2} \right )^{3}+\left ( a^{2}-b^{2} \right )^{3}}{\left ( b-c \right )^{3}+\left ( c-a \right )^{3}+\left ( a-b \right )^{3}}}$ -এর সরলতম মান লিখি।

সমাধানঃ

$\fn_cm \frac{\left ( b^{2}-c^{2} \right )^{3}+\left ( c^{2}-a^{2} \right )^{3}+\left ( a^{2}-b^{2} \right )^{3}}{\left ( b-c \right )^{3}+\left ( c-a \right )^{3}+\left ( a-b \right )^{3}}$

$=\frac{\left ( b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2} \right )\left [\left (b^{2}-c^{2} \right )^{2}+\left (c^{2}-a^{2} \right )^{2}+\left (a^{2}-b^{2} \right )^{2}-\left (b^{2}-c^{2} \right )^{2}\left (c^{2}-a^{2} \right )-\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )-\left (a^{2}-b^{2} \right )\left (b^{2}-c^{2} \right ) \right ]+3\left ( b^{2}-c^{2} \right )\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )}{\left ( b-c+c-a+a-b \right )\left [ \left ( b-c \right )^{2}+\left ( c-a \right )^{2}+\left ( a-b \right )^{2}-\left ( b-c \right )\left ( c-a \right )-\left ( c-a \right )\left ( a-b \right )-\left ( a-b \right ) \left ( b-c \right )\right ]+3\left ( b-c \right )\left ( c-a \right )\left ( a-b \right )}$

${\color{Blue} \left [ \because x^{3}+y^{3}+z^{3}=\left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )+3xyz \right ]}$

$=\frac {3\left ( b^{2}-c^{2} \right )\left (c^{2}-a^{2} \right )\left (a^{2}-b^{2} \right )}{3\left ( b-c \right )\left ( c-a \right )\left ( a-b \right )}$

$=\left ( b+c \right )\left ( c+a \right )\left ( a+b \right )$

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(ii) $\fn_cm {\color{Blue} a^{3}+b^{3}+c^{3}-3abc=0}$ এবং $\fn_cm {\color{Blue} a+b+c\neq 0}$ হলে a, b ও c -এর মধ্যে সম্পর্ক লিখি।

সমাধানঃ

$\fn_cm a^{3}+b^{3}+c^{3}-3abc=0$

বা, $\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0$

$\fn_cm \because a+b+c\neq 0$

$\therefore \left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0$

বা, $2\left ( a^{2}+b^{2}+c^{2}-ab-bc-ca \right )=0\times 2$

বা, $\left ( 2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca \right )=0$

বা, $a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}-2ca+a^{2}=0$

বা, $\left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}=0$

আমরা জানি, দুই বা ততোধিক রাশির বর্গের সমষ্টি শূন্য হলে তারা প্রত্যেকে পৃথক পৃথক ভাবে শূন্য হয়।

$\therefore \left ( a-b \right )^{2}=0$

বা, $a-b=0$

বা, $a=b$

আবার,

$\left ( b-c \right )^{2}=0$

বা, $b-c=0$

বা, $b=c$

আবারও

$\left ( c-a \right )^{2}=0$

বা, $c-a=0$

বা, $c=a$

${\color{DarkGreen} \therefore a=b=c}$ ইহায় a, b ও c -এর মধ্যে সম্পর্ক।

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(iii) $\fn_cm {\color{Blue} a^{2}-b^{2}=224}$ এবং a ও b ঋণাত্মক পূর্ণসংখ্যা হলে (a < b) a ও b -এর মান লিখি।

সমাধানঃ

$\fn_cm a^{2}-b^{2}=224$

বা, $\left ( a+b \right )\left ( a-b \right )=\left ( -14 \right )\times \left ( -16 \right )$

বা, $\left ( a+b \right )\left ( a-b \right )=\left ( -15+1 \right )\times \left ( -15-1 \right )$

উভয়পক্ষ তুলনা করে পাই,

$a=-15,b=-1$

উত্তরঃ নির্ণেয় a ও b এর মান যথাক্রমে -15 ও -1

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(iv) $\fn_cm {\color{Blue} 3x=a+b+c}$ হলে, $\fn_cm {\color{Blue} \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )}$ -এর মান কত লিখি।

সমাধানঃ

$\fn_cm \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )$

ধরি,

x – a = p , x – b = q ও x – c = r

এখন,

p + q + r = x – a + x – b + x – c

বা, p + q + r = 3x – a – b – c

বা, p + q + r = 3x – 3x [ যেহেতু, $\fn_cm {\color{Blue} 3x=a+b+c}$ ]

∴ p + q + r = 0

এখন,

$\fn_cm \left ( x-a \right )^{3}+\left ( x-b \right )^{3}+\left ( x-c \right )^{3}-3\left ( x-a \right )\left ( x-b \right )\left ( x-c \right )$

$=p^{3}+q^{3}+r^{3}-3pqr$

$=\left ( p+q+r \right )\left ( p^{2}+q^{2}+r^{2}-pq-qr-rp \right )+3pqr-3pqr$

$=0\times \left ( p^{2}+q^{2}+r^{2}-pq-qr-rp \right )$

= 0

উৎপাদকে বিশ্লেষণ

3. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

(v) $\fn_cm {\color{Blue} 2x^{2}+px+6=\left ( 2x-a \right )\left ( x-2 \right )}$ একটি অভেদ হলে a ও p -এর মান কত লিখি।

সমাধানঃ

$\fn_cm 2x^{2}+px+6=\left ( 2x-a \right )\left ( x-2 \right )$

বা, $\fn_cm 2x^{2}+px+6=2x^{2}-4x-ax+2a$

বা, $\fn_cm 2x^{2}+px+6=2x^{2}-\left (4+a \right )x+2a$

উভয়পক্ষ তুলনা করে পাই,

$p=-\left ( 4+a \right )$ এবং $6=2a$

$\therefore a=\frac{6}{2}=3$

এবং

$p=-\left ( 4+a \right )$

বা, $p=-\left ( 4+3 \right )$ [ a এর মান বসিয়ে পাই ]

$\therefore p=-7$

উত্তরঃ নির্ণেয় a ও p এর মান যথাক্রমে 3 ও -7

Koshe dekhi 8.5 Class 9

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