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দশম শ্ৰেণীৰ গণিতৰ অনুশীলনী 2.2 -ৰ step by step সম্পূৰ্ণ সমাধান দিব। আপুনি শৈক্ষিক সফলতাৰ বাবে চেষ্টা কৰা শিক্ষাৰ্থী হওঁক বা আপোনাৰ ছাত্ৰ-ছাত্ৰীৰ বাবে ব্যাপক সমাধান বিচৰা শিক্ষক হওঁক, we've got you covered.
ইয়াত, অনুশীলনী 2.2 -ৰ প্ৰতিটো প্ৰশ্নৰ সমাধান লিখি দিয়া হ ল :1. তলৰ দ্বিঘাত বহুপদবোৰৰ শূন্য উলিওৱা আৰু এই শূন্যবোৰ আৰু সহগবোৰৰ মাজৰ সম্পৰ্ক সত্যাপন কৰা।
(i) \(x^2-2x-8\)
উত্তৰ :
\(x^2-2x-8\)
\(=x^2-(4-2)x-8\)
\(=x^2-4x+2x-8\)
\(=x(x-4)+2(x-4)\)
\(=(x+2)(x-4)\)
\(\therefore x+2=0 , x-4=0\)
\(\Rightarrow x=-2 , x=4\)
সত্যাপন :
ইয়াত,
\(a=1\)
\(b=-2\)
\(c=-8\)
\(α=-2\)
\(β=4\)
আমি জানো,
\(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow -2+4=-\frac{-2}{1}\)
\(\Rightarrow 2=\frac{2}{1}\)
\(\Rightarrow 2=2\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(\Rightarrow -2 \times 4 = -\frac{8}{1}\)
\(\Rightarrow -8=-8\)
(ii) \(4s^2-4s+1\)
উত্তৰ :
\(4s^2-4s+1\)
\(=4s^2-(2+2)s+1\)
\(=4s^2-2s-2s+1\)
\(=2s(2s-1)-1(2s-1)\)
\(=(2s-1)(2s-1)\)
\(\therefore 2s-1=0 , 2s-1=0\)
\(\Rightarrow 2s=1 , 2s=1\)
\(\Rightarrow s=\frac{1}{2} , s=\frac{1}{2}\)
সত্যাপন :
ইয়াত,
\(a=4\)
\(b=-4\)
\(c=1\)
\(α=\frac{1}{2}\)
\(β=\frac{1}{2}\)
আমি জানো,
\(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \frac{1}{2}+\frac{1}{2}=-\frac{-4}{4}\)
\(\Rightarrow \frac{1+1}{2}=\frac{4}{4}\)
\(\Rightarrow \frac{2}{2}=1\)
\(\Rightarrow 1=1\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(\Rightarrow \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right)=\frac{1}{4}\)
\(\Rightarrow \frac{1\times1}{2\times2}=\frac{1}{4}\)
\(\Rightarrow \frac{1}{4}=\frac{1}{4}\)
(iii) \(6x^2-3-7x\)
উত্তৰ :
\(6x^2-3-7x\)
\(=6x^2-7x-3\)
\(=6x^2-(9-2)x-3\)
\(=6x^2-9x+2x-3\)
\(=3x(2x-3)+1(2x-3)\)
\(=(3x+1)(2x-3)\)
\(\therefore 3x+1=0 , 2x-3=0\)
\(\Rightarrow 3x=-1 , 2x=3\)
\(\Rightarrow x=-\frac{1}{3} , x=\frac{3}{2}\)
সত্যাপন :
ইয়াত,
\(a=6\)
\(b=-7\)
\(c=-3\)
\(α=-\frac{1}{3}\)
\(β=\frac{3}{2}\)
আমি জানো,
\(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow -\frac{1}{3}+\frac{3}{2}=-\frac{-7}{6}\)
\(\Rightarrow \frac{-2+9}{6}=\frac{7}{6}\)
\(\Rightarrow \frac{7}{6}=\frac{7}{6}\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(\Rightarrow \left(-\frac{1}{3}\right) \times \left(\frac{3}{2}\right)=-\frac{3}{6}\)
\(\Rightarrow \frac{-1\times3}{3\times2}=-\frac{3}{6}\)
\(\Rightarrow -\frac{3}{6}=-\frac{3}{6}\)
(iv) \(4u^2 + 8u\)
উত্তৰ :
\(4u^2 + 8u\)
\(=4u(u + 2)\)
\(\therefore 4u = 0 , u + 2 = 0\)
\(\Rightarrow u = 0 , u = -2\)
সত্যাপন :
ইয়াত,
\(a = 4\)
\(b = 8\)
\(c = 0\)
\(α = 0\)
\(β = -2\)
আমি জানো,
\(\alpha+\beta=\frac{-b}{a}\)
\(⇒0+(-2)=\frac{-8}{4}\)
\(\Rightarrow -2=-2\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(⇒0 \times (-2)=\frac{0}{4}\)
\(\Rightarrow 0=0\)
(v) \(t^2 - 15\)
উত্তৰ :
\(t^2 - 15\)
\( t^2 - (\sqrt{15})^2 \)
অভেদ ব্যৱহাৰ কৰি পাও,
\(= (t + \sqrt{15})(t - \sqrt{15})\)
\(\therefore t + \sqrt{15} = 0 , t - \sqrt{15} = 0\)
\(\Rightarrow t = -\sqrt{15} , t = \sqrt{15}\)
সত্যাপন :
ইয়াত,
\(a = 1\)
\(b = 0\)
\(c = -15\)
\(α = -\sqrt{15}\)
\(β = \sqrt{15}\)
আমি জানো,
\(\alpha+\beta=\frac{-b}{a}\)
\(⇒-\sqrt{15}+\sqrt{15}=\frac{0}{1}\)
\(\Rightarrow 0=0\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(⇒-\sqrt{15} \times \sqrt{15}=\frac{-15}{1}\)
\(\Rightarrow -15=-15\)
(vi) \(3x^2-x-4\)
উত্তৰ :
\(3x^2-x-4\)
\(=3x^2-(4-3)x-4\)
\(=3x^2-4x+3x-4\)
\(=x(3x-4)+1(3x-4)\)
\(=(x+1)(3x-4)\)
\(\therefore x+1=0 , 3x-4=0\)
\(\Rightarrow x=-1 , 3x=4\)
\(\Rightarrow x=-1 , x=\frac{4}{3}\)
সত্যাপন :
ইয়াত,
\(a=3\)
\(b=-1\)
\(c=-4\)
\(α=-1\)
\(β=\frac{4}{3}\)
আমি জানো,
\(\alpha+\beta=-\frac{b}{a}\)
\(⇒(-1)+\frac{4}{3}=-\frac{-1}{3}\)
\(\Rightarrow \frac{-3+4}{3}=\frac{1}{3}\)
\(\Rightarrow \frac{1}{3}=\frac{1}{3}\)
আকৌ,
\(\alpha \times \beta = \frac{c}{a}\)
\(⇒(-1) \times \left(\frac{4}{3}\right)=-\frac{4}{3}\)
\(⇒-\frac{4}{3}=-\frac{4}{3}\)
2. তলৰ যোৰকেইটাৰ সংখ্যা দুটাক ক্ৰমে শূন্যবোৰৰ সমষ্টি আৰু গুণফল হিচাপে ধৰি প্ৰত্যেকৰ ক্ষেত্ৰত একোটা দ্বিঘাত বহুপদ নিৰ্ণয় কৰা ৷
(i) \( \frac{1}{4} \), -1
উত্তৰ :
ধৰো,
\( \alpha + \beta = \frac{1}{4} \)
\( \alpha \times \beta = -1 \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - \left(\frac{1}{4}\right)x + (-1) \)
\( =x^2 - \left(\frac{1}{4}\right)x - 1 \)
যদি, \( k[x^2 - \left(\frac{1}{4}\right)x - 1] \) আৰু \( k=4 \)
∴ \( 4x^2 - 4\times\left(\frac{1}{4}\right)x - 4\times1 \)
\( =4x^2 - x - 4 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( 4x^2 - x - 4 \)
(ii) \( \sqrt{2} \), \( \frac{1}{3} \)
উত্তৰ :
ধৰো,
\( \alpha + \beta = \sqrt{2} \)
\( \alpha \times \beta = \frac{1}{3} \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - (\sqrt{2})x + \frac{1}{3} \)
যদি, \( k[x^2 - (\sqrt{2})x + \frac{1}{3}] \) আৰু \( k=3 \)
∴ \( 3x^2 - 3 \times (\sqrt{2})x + 3 \times \frac{1}{3} \)
\( =3x^2 - 3\sqrt{2}x + 1 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( 3x^2 - 3\sqrt{2}x + 1 \)
(iii) \( 0 \), \( \sqrt{5} \)
উত্তৰ :
ধৰো,
\( \alpha + \beta = 0 \)
\( \alpha \times \beta = \sqrt{5} \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - 0x + \sqrt{5} \)
\( =x^2 + \sqrt{5} \)
যদি, \( k[x^2 + \sqrt{5}] \) আৰু \( k=1 \)
∴ \( 1x^2 + 1 \times \sqrt{5} \)
\( =x^2 + \sqrt{5} \)
∴ দ্বিঘাত বহুপদটো হ'ল \( x^2 + \sqrt{5} \)
(iv) \( 1 \), \( 1 \)
উত্তৰ :
ধৰো,
\( \alpha + \beta = 1 \)
\( \alpha \times \beta = 1 \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - 1x + 1 \)
\( =x^2 - x + 1 \)
যদি, \( k[x^2 - x + 1] \) আৰু \( k=1 \)
∴ \( 1x^2 - 1x + 1\times1 \)
\( =x^2 - x + 1 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( x^2 - x + 1 \)
(v) \( -\frac{1}{4} \), \( \frac{1}{4} \)
উত্তৰ :
ধৰো,
\( \alpha + \beta = -\frac{1}{4} \)
\( \alpha \times \beta = \frac{1}{4} \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4} \)
\( =x^2 + \frac{1}{4}x + \frac{1}{4} \)
যদি, \( k[x^2 + \frac{1}{4}x + \frac{1}{4}] \) আৰু \( k=4 \)
∴ \( 4x^2 + 4\times\left(\frac{1}{4}\right)x + 4\times\frac{1}{4} \)
\( =4x^2 + x + 1 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( 4x^2 + x + 1 \)
(vi) \( 4 \), \( 1 \)
উত্তৰ :
ধৰো,
\( \alpha + \beta = 4 \)
\( \alpha \times \beta = 1 \)
আমি জানো,
\( x^2 - (\alpha + \beta)x + \alpha \times \beta \)
\( =x^2 - 4x + 1 \)
যদি, \( k[x^2 - 4x + 1] \) আৰু \( k=1 \)
∴ \( 1x^2 - 4x + 1\times1 \)
\( =x^2 - 4x + 1 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( x^2 - 4x + 1 \)
3. দ্বিঘাত বহুপদবোৰ নিৰ্ণয় কৰা যাৰ শূন্যকেইটা তলত দিয়া ধৰণৰ :
(i) \( -4 \), \( \frac{3}{2} \)
উত্তৰ :
ধৰো,
\( \alpha = -4 \)
\( \beta = \frac{3}{2} \)
∵ \( \alpha + \beta = -\frac{b}{a} \)
\( \Rightarrow -4 + \frac{3}{2} = -\frac{b}{a} \)
\( \Rightarrow \frac{-8+3}{2} = -\frac{b}{a} \)
\( \Rightarrow -\frac{5}{2} = -\frac{b}{a} \)
ইয়াত,
\( -b = -5 \)
\( \Rightarrow b = 5 →(i)\)
\( a = 2 →(ii)\)
আকৌ,
\( \alpha\times \beta = \frac{c}{a} \)
\( \Rightarrow -4 \times \frac{3}{2} = \frac{c}{a} \)
\( ⇒-4 \times \frac{3}{2} = \frac{c}{2} [∵ a=2]\)
\( \Rightarrow -\frac{12}{2} = \frac{c}{2} \)
\( \Rightarrow 2c = -12 \times 2 \)
\( \Rightarrow 2c = -24 \)
\( \Rightarrow c = -\frac{24}{2} \)
\( \Rightarrow c = -12 →(iii) \)
∵ \( ax^2 + bx + c = 0 \)
∴ (i), (ii) আৰু (iii) ৰ পৰা,
∴ \( 2x^2 + 5x + (-12) = 0 \)
\( \Rightarrow 2x^2 + 5x - 12 = 0 \)
∴ দ্বিঘাত বহুপদটো হ'ল \( 2x^2 + 5x - 12 = 0 \)
(ii) 5 আৰু 2
উত্তৰ :
ধৰো,
\( \alpha = 5 \)
\( \beta = 2 \)
∵ \( \alpha + \beta = -\frac{b}{a} \)
\(⇒ 5 + 2 = -\frac{b}{a} \)
\(⇒ 7 = -\frac{b}{a} \)
ইয়াত,
\( -b = 7 \)
\( \Rightarrow b = -7 \) →(i)
\( a = 1 \) →(ii)
আকৌ,
\( \alpha \times \beta = \frac{c}{a} \)
\(⇒ 5 \times 2 = \frac{c}{a} \)
\(⇒ 10 = \frac{c}{1} [∵ a=1] \)
\(⇒ c = 10 \) →(iii)
∵ \( ax^2 + bx + c = 0 \)
∴ (i), (ii) আৰু (iii) ৰ পৰা,
∴ \( 1x^2 + (-7x) + 10 = 0 \)
\( \Rightarrow x^2 - 7x + 10 = 0 \)
∴ দ্বিঘাত বহুপদটো হ'ল = \( x^2 - 7x + 10 = 0 \)
(iii) \( \frac{1}{3} \) আৰু -1
উত্তৰ :
ধৰো,
\( \alpha = \frac{1}{3} \)
\( \beta = -1 \)
∵ \( \alpha + \beta = -\frac{b}{a} \)
\(⇒ \frac{1}{3} +(- 1 )= -\frac{b}{a} \)
\(⇒ \frac{1}{3} - 1 = -\frac{b}{a} \)
\( \Rightarrow \frac{1-3}{3} = -\frac{b}{a} \)
\( ⇒-\frac{2}{3} = -\frac{b}{a} \)
ইয়াত,
\( -b = -2 \)
\( \Rightarrow b =2 \) →(i)
\( a = 3 \) →(ii)
আকৌ,
\( \alpha \times \beta = \frac{c}{a} \)
\(⇒ \frac{1}{3} \times (-1) = \frac{c}{a} \)
\(⇒ -\frac{1}{3} = \frac{c}{3} \) [∵ a=3]
\(⇒ -3c = 3 \)
\( ⇒c = -\frac{3}{3} \)
\(⇒ c = -1 \) →(iii)
∵ \( ax^2 + bx + c = 0 \)
∴ (i), (ii) আৰু (iii) ৰ পৰা,
∴ \( 3x^2 + 2x+(-1) = 0 \)
⇒ \( 3x^2 + 2x-1 = 0 \)
∴ দ্বিঘাত বহুপদটো হ'ল = \( 3x^2 + 2x-1 = 0 \)
(iv) \( \frac{3}{2} \) আৰু -2
উত্তৰ :
ধৰো,
\( \alpha = \frac{3}{2} \)
\( \beta = -2 \)
∵ \( \alpha + \beta = -\frac{b}{a} \)
\(⇒ \frac{3}{2} +(- 2 )= -\frac{b}{a} \)
\(⇒ \frac{3}{2} - 2 = -\frac{b}{a} \)
\( \Rightarrow \frac{3-4}{2} = -\frac{b}{a} \)
\( ⇒-\frac{1}{2} = -\frac{b}{a} \)
ইয়াত,
\( -b = -1 \)
\( \Rightarrow b =1 \) →(i)
\( a = 2 \) →(ii)
আকৌ,
\( \alpha \times \beta = \frac{c}{a} \)
\(⇒ \frac{3}{2} \times (-2) = \frac{c}{a} \)
\( ⇒-\frac{6}{2}= \frac{c}{2} \) [∵ a=2]
\( ⇒-3= \frac{c}{2} \)
\(⇒ c = -6 \) →(iii)
∵ \( ax^2 + bx + c = 0 \)
(i), (ii) আৰু (iii) ৰ পৰা,
∴ \( 2x^2 + x + (-6) = 0 \)
⇒ \( 2x^2 + x -6 = 0 \)
∴ দ্বিঘাত বহুপদটো হ'ল = \( 2x^2 + x -6 = 0 \)
Remember, with the right guidance and practice makes perfect in mathematics. Keep exploring, experimenting, and solving problems to excel in this fascinating subject.

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