Class eight math solution

অষ্টম শ্রেণী
অনুশীলনী-৪.১

(ক) সমাধান :
\left( {5a + 7b} \right) রাশির বর্গ = {\left( {5a + 7b} \right)^2}
= {\left( {5a} \right)^2} + 2 \times 5a \times 7b + {\left( {7b} \right)^2}
= 25{a^2} + 70ab + 49{b^2} (Ans.)

(খ) সমাধান :
\left( {6x + 3} \right) রাশির বর্গ = {\left( {6x + 3} \right)^2}
= {\left( {6x} \right)^2} + 2 \times 6x \times 3 + {\left( 3 \right)^2}
= 36{x^2} + 36x + {9} (Ans.)

(গ) সমাধান :
\left( {7p - 2q} \right) রাশির বর্গ = {\left( {7p - 2q} \right)^2}
= {\left( {7p} \right)^2} - 2 \times 7p \times 2q + {\left( {2q} \right)^2}
= 49{p^2} - 28pq + 4{q^2}  (Ans.)

(ঘ) সমাধান :
\left( {ax - by} \right) রাশির বর্গ = {\left( {ax - by} \right)^2}
= {\left( {ax} \right)^2} - 2 \times ax \times by + {\left( {by} \right)^2}
= {a^2}{x^2} - 2abxy + {b^2}{y^2}  (Ans.)

(ঙ) সমাধান :
\left( {{x^3} + xy} \right) রাশির বর্গ = {\left( {{x^3} + xy} \right)^2}
= {\left( {{x^3}} \right)^2} + 2 \times {x^3} \times xy + {\left( {xy} \right)^2}
= {x^6} + 2{x^4}y + {x^2}{y^2}  (Ans.)

(চ) সমাধান :
\left( {11a - 12b} \right) রাশির বর্গ = {\left( {11a - 12b} \right)^2}
= {\left( {11a} \right)^2} - 2 \times 11a \times 12b + {\left( {12b} \right)^2}
= 121{a^2} - 264ab + 144{b^2} (Ans.)

(ছ) সমাধান :
\left( {6{x^2}y - 5x{y^2}} \right) রাশির বর্গ = {\left( {6{x^2}y - 5x{y^2}} \right)^2}
= {\left( {6{x^2}y} \right)^2} - 2 \times 6{x^2}y \times 5x{y^2} + {\left( {5x{y^2}} \right)^2}
= 36{x^4}{y^2} - 60{x^3}{y^3} + 25{x^2}{y^4} (Ans.)

(জ) সমাধান :
\left( { - x - y} \right) রাশির বর্গ = {\left( { - x - y} \right)^2}
= {\left( { - x - y} \right)^2}
= {\left\{ {\left( { - x} \right) + ( - y)} \right\}^2}
= {\left( { - x} \right)^2} + 2 \times \left( { - x} \right) \times \left( { - y} \right) + {\left( { - y} \right)^2}
= {x^2} + 2xy + {y^2} (Ans.)

(ঝ) সমাধান :
\left( { - xyz - abc} \right)  রাশির বর্গ = {\left( { - xyz - abc} \right)^2}
= {\left\{ {\left( { - 1} \right)\left( {xyz + abc} \right)} \right\}^2}
= {\left( {xyz + abc} \right)^2}
= {\left( {xyz} \right)^2} + 2 \times xyz \times abc + {\left( {abc} \right)^2}
= {x^2}{y^2}{z^2} + 2abcxyz + {a^2}{b^2}{c^2}  (Ans)

(ঞ) সমাধান :
\left( {{a^2}{x^3} - {b^2}{y^4}} \right)  রাশির বর্গ = {\left( {{a^2}{x^3} - {b^2}{y^4}} \right)^2}
= {\left( {{a^2}{x^3}} \right)^2} - 2 \times {a^2}{x^3} \times {b^2}{y^4} + {\left( {{b^2}{y^4}} \right)^2}
= {a^4}{x^6} - 2{a^2}{x^3}{b^2}{y^4} + {b^4}{y^8} (Ans)

(ট) সমাধান :
\left( {108} \right)  সংখ্যাটির বর্গ = {\left( {100 + 8} \right)^2}
= {\left( {100} \right)^2} + 2 \times 100 \times 8 + {\left( 8 \right)^2}
= 10000 + 1600 + 64
= 11664 (Ans)

(ঠ) সমাধান :
606  সংখ্যাটির বর্গ = = {\left( {600 + 6} \right)^2}
= {\left( {600} \right)^2} + 2 \times 600 \times 6 + {\left( 6 \right)^2}
= 360000 + 7200 + 36
= 367236

(ড) সমাধান:
\left( {597} \right) সংখ্যাটির বর্গ: = {\left( {600 - 3} \right)^2}
= {\left( {600} \right)^2} - 2 \times 600 \times 3 + {\left( 3 \right)^2}
= 360000 - 3600 + 9
= 356409

সমাধান (ঢ):
a - b + c রাশির বর্গ ={(a - b + c)^2}
= {\{ a - (b - c)\} ^2}
= {a^2} - 2a(b - c) + {(b - c)^2}
= {a^2} - 2ab + 2ac + {b^2} - 2bc + {c^2}
= {a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ca  (Ans.)
সমাধান (ণ):
ax + b + 2 রাশির বর্গ = {(ax + b + 2)^2}
= {\{ ax + (b + 2)\} ^2}
= {(ax)^2} + 2.ax.(b + 2) + {(b + 2)^2}
= {a^2}{x^2} + 2axb + 4ax + {b^2} + 2.b.2 + {2^2}
= {a^2}{x^2} + {b^2} + 2abx + 4b + 4ax + 4  (Ans.)

সমাধান (ত):
xy + yz - zx রাশির বর্গ {(xy + yz - zx)^2}
= {\{ xy + (yz - zx)\} ^2}
= {(xy)^2} + 2 \times xy \times (yz - zx) + {(yz - zx)^2}
= {x^2}{y^2} + 2x{y^2}z - 2{x^2}yz + {(yz)^2} - 2 \times yz \times zx + {(zx)^2}
= {x^2}{y^2} + 2x{y^2}z - 2{x^2}yz + {y^2}{z^2} - 2y{z^2}x + {z^2}{x^2}
= {x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} + 2x{y^2}z - 2{x^2}yz - 2y{z^2}x  (Ans.)

সমাধান (থ)
3p + 2q - 5r রাশির বর্গ = {(3p + 2q - 5r)^2}
= {\{ 3p + (2q - 5r)\} ^2}
= {(3p)^2} + 2 \times 3p \times (2q - 5r) + {(2q - 5r)^2}
= 9{p^2} + 6p \times (2q - 5r) + {(2q)^2} - 2 \times 2q \times 5r + {(5r)^2}
= 9{p^2} + 12pq - 30pr + 4{q^2} - 20qr + 25{r^2}
= 9{p^2} + 4{q^2} + 25{r^2} + 12pq - 30pr - 20qr   (Ans.)

সমাধান (দ)
{x^2} - {y^2} - {z^2} রাশির বর্গ = {({x^2} - {y^2} - {z^2})^2}
= {\{ {x^2} - ({y^2} + {z^2})\} ^2}
= {({x^2})^2} - 2 \times {x^2} \times ({y^2} + {z^2}) + {({y^2} + {z^2})^2}
= {x^4} - 2{x^2}{y^2} - 2{x^2}{z^2} + {({y^2})^2} + 2 \times {y^2} \times {z^2} + {({z^2})^2}
= {x^4} - 2{x^2}{y^2} - 2{x^2}{z^2} + {y^4} + 2{y^2}{z^2} + {z^4}
= {x^4} + {y^4} + {z^4} - 2{x^2}{y^2} + 2{y^2}{z^2} - 2{x^2}{z^2}

সমাধান (ধ):
7{a^2} + 8{b^2} - 5{c^2} রাশির বর্গ = {(7{a^2} + 8{b^2} - 5{c^2})^2}
= {\{ 7{a^2} + (8{b^2} - 5{c^2})\} ^2}
= {(7{a^2})^2} + 2 \times 7{a^2} \times (8{b^2} - 5{c^2}) + {(8{b^2} - 5{c^2})^2}
= 49{a^4} + 14{a^2}(8{b^2} - 5{c^2}) + {(8{b^2})^2} - 2 \times 8{b^2} \times 5{c^2} + {(5{c^2})^2}
= 49{a^4} + 112{a^2}{b^2} - 70{a^2}{c^2} + 64{b^4} - 80{b^2}{c^2} + 25{c^4}

২। সরল কর:
(ক){(x + y)^2} + 2(x + y)(x - y) + {(x - y)^2}
(খ){(2a + 3b)^2} - 2(2a + 3b)(3b - a) + {(3b - a)^2}
(গ){(3{x^2} + 7{y^2})^2} + 2(3{x^2} + 7{y^2})(3{x^2} - 7{y^2}) + (3{x^2} - 7{y^2})
(ঘ){(8x + y)^2} - (16x + 2y)(5x + y) + {(5x + y)^2}
(ঙ){(5{x^2} - 3x - 2)^2} + {(2 + 5{x^2} - 3{x^2})^2} - 2(5{x^2} - 3x - 2)(2 + 5{x^2} - 3x)

সমাধান (ক) :
মনেকরি,
x + y = a এবং x - y = b
\therefore প্রদত্ত রাশি = {(a)^2} + 2ab + {(b)^2}
= {a^2} + 2ab + {b^2}
= {(a + b)^2}
= {(x + y + x - y)^2}
= {(2x)^2}
= 4{x^2}  (Ans.)

সমাধান (খ):
মনেকরি,
2a + 3b = x এবং 3b - a = y
\thereforeপ্রদত্ত রাশি = {(x)^2} - 2.x.y + {(y)^2}
= {x^2} - 2xy + {y^2}
= {(x - y)^2}
= {\{ (2a + 3b) - (3b - a)\} ^2}
= {(2a + 3b - 3b + a)^2}
= {(3a)^2}
= 9{a^2}  (Ans.)

সমাধান (গ):
মনেকরি, 3{x^2} + 7{y^2} = a এবং 3{x^2} - 7{y^2} = b
\therefore প্রদত্ত রাশি = {(a)^2} + 2.a.b + {(b)^2}
= {a^2} + 2ab + {b^2}
= {(a + b)^2}
= {(3{x^2} + 7{y^2} + 3{x^2} - 7{y^2})^2}
= {(6{x^2})^2}
= 36{x^4}  (Ans.)

সমাধান (ঘ):
প্রদত্ত রাশি= {(8x + y)^2} - (16x + 2y)(5x + y) + {(5x + y)^2}
= {(8x + y)^2} - 2(8x + y)(5x + y) + {(5x + y)^2}
মনেকরি,
8x + y = a এবং 5x + y = b
\therefore প্রদত্ত রাশি= {(a)^2} - 2(a)(b) + {(b)^2}
= {a^2} - 2ab + {b^2}
= {(a - b)^2}
= {\{ (8x + y) - (5x + y)\} ^2}
= {(8x + y - 5x - y)^2}
= {(3x)^2}
= 9{x^2}  (Ans.)

সমাধান (ঙ):
মনেকরি, 5{x^2} - 3x - 2 = a  এবং 2 + 5{x^2} - 3x = b
\therefore প্রদত্ত রাশি = {(a)^2} + {(b)^2} - 2(a)(b)
= {a^2} + {b^2} - 2ab
= {a^2} - 2ab + {b^2}
= {(a - b)^2}
= {\{ (5{x^2} - 3x - 2) - (2 + 5{x^2} - 3x)\} ^2}
= {(5{x^2} - 3x - 2 - 2 - 5{x^2} + 3x)^2}
= {( - 4)^2}
= 16  (Ans.)

৩। সূত্র প্রয়োগ করে গুণফল নির্ণয় কর:
(ক) (x + 7)(x - 7)                (খ) (5x + 13)(5x - 13)
(গ) (xy + yz)(xy - yz)           (ঘ) (ax + b)(ax - b)
(ঙ) (a + 3)(a + 4)                (চ) (ax + 3)(ax + 4)
(ছ) (6x + 17)(6x - 13)          (জ)({a^2} + {b^2})({a^2} - {b^2})({a^4} + {b^4})
(ঝ) (ax - by + cz)(ax + by - cz) (ঞ) (3a - 10)(3a - 5) 
(ট) (5a + 2b - 3c)(5a + 2b + 3c)      (ঠ) (ax + by + 5)(ax + by + 3)

সমাধান (ক):
প্রদত্ত রাশি = (x + 7)(x - 7)
= {x^2} - {7^2}  [\because (a + b)(a - b) = {a^2} - {b^2}]
= {x^2} - 49  (Ans.)

সমাধান (খ):
প্রদত্ত রাশি= (5x + 13)(5x - 13)
= {(5x)^2} - {(13)^2} [\because (a + b)(a - b) = {a^2} - {b^2}]
= 25{x^2} - 169 (Ans.)

সমাধান (গ):
প্রদত্ত রাশি = (xy + yz)(xy - yz)
= {(xy)^2} - {(yz)^2} [\because (a + b)(a - b) = {a^2} - {b^2}]
= {x^2}{y^2} - {y^2}{z^2}  (Ans.)

সমাধান (ঘ):
প্রদত্ত রাশি = (ax + b)(ax - b)
= {(ax)^2} - {b^2}  [\because (a + b)(a - b) = {a^2} - {b^2}]
= {a^2}{x^2} - {b^2} (Ans.)

সমাধান (ঙ):
প্রদত্ত রাশি= (a + 3)(a + 4)
= {a^2} + (3 + 4)a + 3 \times 4   [\because (x + a)(x + b) = {x^2} + (a + b)x + ab]
= {a^2} + 7a + 12 (Ans.)

সমাধান (চ):
প্রদত্ত রাশি = (ax + 3)(ax + 4)
= {(ax)^2} + (3 + 4)ax + 3 \times 4  [\because (x + a)(x + b) = {x^2} + (a + b)x + ab]
= {a^2}{x^2} + 7ax + 12  (Ans.)

সমাধান (ছ):
প্রদত্ত রাশি = (6x + 17)(6x - 13)
= (6x + 17)\{ 6x + ( - 13)\}
= {(6x)^2} + \{ 17 + ( - 13)\} 6x + (17)( - 13)
= {(6x)^2} + (17 - 13)6x - 221
= 36{x^2} + 4 \times 6x - 221
= 36{x^2} + 24x - 221

সমাধান (জ)
প্রদত্ত রাশি ({a^2} + {b^2})({a^2} - {b^2})({a^4} + {b^4})
= \{ {({a^2})^2} - {({b^2})^2}\} ({a^4} + {b^4})
= ({a^4} - {b^4})({a^4} + {b^4})
= {({a^4})^2} - {({b^4})^2}
= {a^8} - {b^8}  (Ans.)

সমাধান (ঝ)
প্রদত্ত রাশি= (ax - by + cz)(ax + by - cz)
= \{ ax - (by - cz)\} \{ ax + (by - cz)\}
= {(ax)^2} - {(by - cz)^2}[\because (a + b)(a - b) = {a^2} - {b^2}]
= {a^2}{x^2} - \{ {(by)^2} - 2.by.cz + {(cz)^2}\}
= {a^2}{x^2} - ({b^2}{y^2} - 2bcyz + {c^2}{z^2})
= {a^2}{x^2} - {b^2}{y^2} + 2bcyz - {c^2}{z^2}
= {a^2}{x^2} - {b^2}{y^2} - {c^2}{z^2} + 2bcyz  (Ans.)

সমাধান (ঞ):
প্রদত্ত রাশি= (3a - 10)(3a - 5)
= \{ 3a + ( - 10)\} \{ 3a + ( - 5)\}
= {(3a)^2} + ( - 10 - 5)3a + ( - 10)( - 5)
= 9{a^2} + ( - 15)3a + ( - 50)
= 9{a^2} - 45a - 50  (Ans.)

সমাধান (ট):
প্রদত্ত রাশি=(5a + 2b - 3c)(5a + 2b + 3c)
= \{ (5a + 2b) - 3c\} \{ (5a + 2b) + 3c\}
= {(5a + 2b)^2} - {(3c)^2}
= {(5a)^2} + 2.5a.2b + {(2b)^2} - {(3c)^2}
= 25{a^2} + 20ab + 4{b^2} - 9{c^2}
= 25{a^2} + 4{b^2} - 9{c^2} + 20ab (Ans.)

সমাধান (ঠ):
প্রদত্ত রাশি = (ax + by + 5)(ax + by + 3)
= \{ (ax + by) + 5\} \{ (ax + by) + 3\}
= {(ax + by)^2} + (5 + 3)(ax + by) + 5 \times 3
= {(ax)^2} + 2 \times ax \times by + {(by)^2} + 8(ax + by) + 15
= {a^2}{x^2} + 2abxy + {b^2}{y^2} + 8ax + 8by + 15
= {a^2}{x^2} + {b^2}{y^2} + 8ax + 8by + 2abxy + 15  (Ans.)

৪। a = 4, b = 6 এবং c = 3 হলে 4{a^2}{b^2} - 16a{b^2}c + 16{b^2}{c^2} এর মান নির্ণয় কর।
সমাধান:
দেওয়া আছে,
a = 4, b = 6 এবং c = 3
প্রদত্ত রাশি,
4{a^2}{b^2} - 16a{b^2}c + 16{b^2}{c^2}
= {(2ab)^2} - 2 \times 2ab \times 4bc + {(4bc)^2}
= {(2ab - 4bc)^2}   [\because {(a - b)^2} = {a^2} - 2ab + {b^2})]
= {(2 \times 4 \times 6 - 4 \times 6 \times 3)^2}
= {(48 - 72)^2}
= {( - 24)^2}
= 576   (Ans.)

৫। x - \frac{1}{x} = 3 হলে, {x^2} + \frac{1}{{{x^2}}}  এর মান নির্ণয় কর।
সমাধান:
দেওয়া আছে, x - \frac{1}{x} = 3
\therefore প্রদত্ত রাশি = {x^2} + \frac{1}{{{x^2}}}
= {(x)^2} + {\left( {\frac{1}{x}} \right)^2}
= {\left( {x - \frac{1}{x}} \right)^2} + 2 \times x \times \frac{1}{x}
[\because {a^2} + {b^2} = {(a - b)^2} + 2ab]
= {(3)^2} + 2
= 9 + 2
= 11

৬। a + \frac{1}{a} = 4 হলে, {a^4} + \frac{1}{{{a^4}}} এর মান কত? সমাধান:
দেওয়া আছে, a + \frac{1}{a} = 4
\therefore প্রদত্ত রাশি= {a^4} + \frac{1}{{{a^4}}}
= {({a^2})^2} + {\left( {\frac{1}{{{a^2}}}} \right)^2}
= {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} - 2 \times {a^2} \times \frac{1}{{{a^2}}}   [\because {a^2} + {b^2} = {(a + b)^2} - 2ab]
= {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} - 2
= {\left\{ {{{(a)}^2} + {{\left( {\frac{1}{a}} \right)}^2}} \right\}^2} - 2
= {\left\{ {{{\left( {a + \frac{1}{a}} \right)}^2} - 2 \times a \times \frac{1}{a}} \right\}^2} - 2
= {\left\{ {{{\left( {a + \frac{1}{a}} \right)}^2} - 2} \right\}^2} - 2  [\because {a^2} + {b^2} = {(a + b)^2} - 2ab]
= {\{ {(4)^2} - 2\} ^2} - 2
= {\left( {16 - 2} \right)^2} - 2
= {(14)^2} - 2
= 196 - 2
= 194   (Ans.)

৭। m = 6, n = 7 হলে, 16({m^2} + {n^2}) + 56({m^2} + {n^2})(3{m^2} - 2{n^2}) + 49{(3{m^2} - 2{n^2})^2} এর মান নির্ণয় কর।
সমাধান:
দেওয়া আছে,
m = 6, n = 7
মনেকরি, {m^2} + {n^2} = a এবং 3{m^2} - 2{n^2} = b প্রদত্ত রাশি,
16({m^2} + {n^2}) + 56({m^2} + {n^2})(3{m^2} - 2{n^2}) + 49{(3{m^2} - 2{n^2})^2}
= 16{a^2} + 56ab + 49{b^2}     [মান বসিয়ে]
= {(4a)^2} + 2 \times 4a \times 7b + {(7b)^2}
= {(4a + 7b)^2}
= {\{ 4({m^2} + {n^2}) + 7(3{m^2} - 2{n^2})\} ^2}
= {(4{m^2} + 4{n^2} + 21{m^2} - 14{n^2})^2}
= {(25{m^2} - 10{n^2})^2}
= {\{ 25 \times {(6)^2} - 10 \times {(7)^2}\} ^2}
= {(25 \times 36 - 10 \times 49)^2}
= {(900 - 490)^2}
= {(410)^2}
= 168100   (Ans.)

৮। a - \frac{1}{a} = m হলে, দেখাও যে, {a^4} + \frac{1}{{{a^4}}} = {m^4} + 4{m^2} + 2
সমাধান:
দেওয়া আছে,
a - \frac{1}{a} = m
এখন,
{a^4} + \frac{1}{{{a^4}}}
= {({a^2})^2} + {\left( {\frac{1}{{{a^2}}}} \right)^2}
= {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} - 2 \times {a^2} \times \frac{1}{{{a^2}}}    [\because {a^2} + {b^2} = {(a + b)^2} - 2ab]
= {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} - 2
= {\left\{ {{{\left( {a - \frac{1}{a}} \right)}^2} + 2 \times a \times \frac{1}{a}} \right\}^2} - 2 [\because {a^2} + {b^2} = {(a - b)^2} + 2ab]
= {\left\{ {{{\left( {a - \frac{1}{a}} \right)}^2} + 2} \right\}^2} - 2 
= {\left\{ {{{(m)}^2} + 2} \right\}^2} - 2
= {({m^2} + 2)^2} - 2
= {({m^2})^2} + 2 \times {m^2} \times 2 + {2^2} - 2
= {m^4} + 4{m^2} + 4 - 2
= {m^4} + 4{m^2} + 2
\therefore {a^4} + \frac{1}{{{a^4}}} = {m^4} + 4{m^2} + 2  (দেখানো হলো)

৯। x - \frac{1}{x} = 4 হলে, প্রমাণ কর যে, {x^2} + {\left( {\frac{1}{x}} \right)^2} = 18
সমাধান:
দেওয়া আছে, x - \frac{1}{x} = 4
প্রমান করতে হবে যে, {x^2} + {\left( {\frac{1}{x}} \right)^2} = 18
এখন,
{x^2} + {\left( {\frac{1}{x}} \right)^2}
= {\left( {x - \frac{1}{x}} \right)^2} + 2 \times x \times \frac{1}{x}   [\because {a^2} + {b^2} = {(a - b)^2} + 2ab]
= {\left( {x - \frac{1}{x}} \right)^2} + 2
= {(4)^2} + 2
= 16 + 2
= 18

\therefore {x^2} + {\left( {\frac{1}{x}} \right)^2} = 18  (প্রমাণিত)

১০। m + \frac{1}{m} = 2 হলে, প্রমান কর যে, {m^4} + \frac{1}{{{m^4}}} = 2

সমাধান:
দেওয়া আছে,

m + \frac{1}{m} = 2
বা, {\left( {m + \frac{1}{m}} \right)^2} = {2^2}   [উভয় পার্শ্বে বর্গ করে]
বা, {m^2} + 2 \times m \times \frac{1}{m} + {\left( {\frac{1}{m}} \right)^2} = 4 [\because {(a + b)^2} = {a^2} + 2ab + {b^2}]
বা, {m^2} + 2 + \frac{1}{{{m^2}}} = 4
বা, {m^2} + \frac{1}{{{m^2}}} = 4 - 2
বা, {m^2} + \frac{1}{{{m^2}}} = 2
বা, {\left( {{m^2} + \frac{1}{{{m^2}}}} \right)^2} = {2^2}   [উভয় পার্শ্বে বর্গ করে]
বা, {({m^2})^2} + 2 \times {m^2} \times \frac{1}{{{m^2}}} + {\left( {\frac{1}{{{m^2}}}} \right)^2} = 4 [\because {(a + b)^2} = {a^2} + 2ab + {b^2}]
বা, {m^4} + 2 + \frac{1}{{{m^4}}} = 4
বা, {m^4} + \frac{1}{{{m^4}}} = 4 - 2
বা, {m^4} + \frac{1}{{{m^4}}} = 2

\therefore {m^4} + \frac{1}{{{m^4}}} = 2  (প্রমাণিত)

১১। x + y = 12 এবং xy = 27 হলে, {(x - y)^2} এবং {x^2} + {y^2} এর মান নির্ণয় কর।
সমাধান:
দেওয়া আছে,
x + y = 12 এবং xy = 27

এখন,
{(x - y)^2}
= {(x + y)^2} - 4xy
= {(12)^2} - 4 \times 27
= 144 - 108
= 36   (Ans.)

১২। a + b = 13 এবং a - b = 3 হলে, 2{a^2} + 2{b^2} এবং ab এর মান নির্ণয় কর।
সমাধান:
দেওয়া আছে,
a + b = 13 এবং a - b = 3

এখন,2{a^2} + 2{b^2}
= 2({a^2} + {b^2})
= {(a + b)^2} + {(a - b)^2}
= {(13)^2} + {(3)^2}
= 169 + 9
= 178  (Ans.)

আবার,
ab = {\left( {\frac{{a + b}}{2}} \right)^2} - {\left( {\frac{{a - b}}{2}} \right)^2}
= {\left( {\frac{{13}}{2}} \right)^2} - {\left( {\frac{3}{2}} \right)^2}
= \frac{{169}}{4} - \frac{9}{4}
= \frac{{169 - 9}}{4}
= \frac{{160}}{4}
= 40    (Ans.)

১৩। দুইটি রাশির বর্গের অন্তররূপে প্রকাশ কর:
(ক)(5p - 3q)(p + 7q)                 (খ)(6a + 9b)(7b - 8a)
(গ)(3x + 5y)(7x - 5y)                (ঘ)(5x + 13)(5x - 13)
সমাধান (ক) :
আমরা জানি, ab = {\left( {\frac{{a + b}}{2}} \right)^2} - {\left( {\frac{{a - b}}{2}} \right)^2}
\therefore (5p - 3q)(p + 7q)
= {\left\{ {\frac{{(5p - 3q) + (p + 7q)}}{2}} \right\}^2} - {\left\{ {\frac{{(5p - 3q) - (p + 7q)}}{2}} \right\}^2}
= {\left( {\frac{{5p - 3q + p + 7q}}{2}} \right)^2} - {\left( {\frac{{5p - 3q - p - 7q}}{2}} \right)^2}
= {\left( {\frac{{6p + 4q}}{2}} \right)^2} - {\left( {\frac{{4p - 10q}}{2}} \right)^2}
= {\left\{ {\frac{{2(3p + 2q)}}{2}} \right\}^2} - {\left\{ {\frac{{2(2p - 5q)}}{2}} \right\}^2}
= {(3p + 2q)^2} - {(2p - 5q)^2}    (Ans.)

সমাধান (খ):
আমরা জানি, ab = {\left( {\frac{{a + b}}{2}} \right)^2} - {\left( {\frac{{a - b}}{2}} \right)^2}
\therefore (6a + 9b)(7b - 8a)
= {\left\{ {\frac{{(6a + 9b) + (7b - 8a)}}{2}} \right\}^2} - {\left\{ {\frac{{(6a + 9b) - (7b - 8a)}}{2}} \right\}^2}
= {\left( {\frac{{6a + 9b + 7b - 8a}}{2}} \right)^2} - {\left( {\frac{{6a + 9b - 7b + 8a}}{2}} \right)^2}
= {\left( {\frac{{ - 2a + 16b}}{2}} \right)^2} - {\left( {\frac{{14a + 2b}}{2}} \right)^2}
= {\left\{ {\frac{{2( - a + 8b)}}{2}} \right\}^2} - {\left\{ {\frac{{2(7a + b)}}{2}} \right\}^2}
= {( - a + 8b)^2} - {(7a + b)^2}
= {(8b - a)^2} - {(b + 7a)^2}   (Ans.)
বি.দ্র: {( - a + 8b)^2} = {(8b - a)^2} একই কথা

সমাধান (গ) :
আমরা জানি, ab = {\left( {\frac{{a + b}}{2}} \right)^2} - {\left( {\frac{{a - b}}{2}} \right)^2}
\therefore (3x + 5y)(7x - 5y)
= {\left\{ {\frac{{(3x + 5y) + (7x - 5y)}}{2}} \right\}^2} - {\left\{ {\frac{{(3x + 5y) - (7x - 5y)}}{2}} \right\}^2}
= {\left( {\frac{{3x + 5y + 7x - 5y}}{2}} \right)^2} - {\left( {\frac{{3x + 5y - 7x + 5y}}{2}} \right)^2}
= {\left( {\frac{{10x}}{2}} \right)^2} - {\left( {\frac{{10y - 4x}}{2}} \right)^2}
= {\left( {\frac{{2 \times 5x}}{2}} \right)^2} - {\left( {\frac{{2(5y - 2x)}}{2}} \right)^2}
= {(5x)^2} - {(5y - 2x)^2}
= {(5x)^2} - {(2x - 5y)^2}   (Ans.)

সমাধান (ঘ):
আমরা জানি, (a + b)(a - b) = {a^2} - {b^2}
\therefore (5x + 13)(5x - 13)
= {(5x)^2} - {(13)^2}    (Ans.)

১৪। দুইটি সংখ্যা a ও b যেখানে a > b । সংখ্যাদ্বয়ের যোগফল 12 এবং গুণফল 32
(ক) সূত্রের সাহায্যে গুণ করো: (2x + 3)(2x - 7)
(খ) 2{a^2} + 2{b^2} এর মান নির্ণয় করো।
(গ) প্রমাণ কর যে, {(a + 2b)^2} - 5{b^2} = 176

সমাধান (ক) :
আমরা জানি,
(x + a)(x + b) = {x^2} + (a + b)x + ab
\therefore (2x + 3)(2x - 7)
= (2x + 3)\{ (2x + ( - 7)\}
= {(2x)^2} + \{ 3 + ( - 7)\} 2x + 3 \times ( - 7)
= {(2x)^2} + (3 - 7)2x - 21
= 4{x^2} + ( - 4)2x - 21
= 4{x^2} - 8x - 21   (Ans.)

সমাধান (খ):
দেওয়া আছে,
a + b = 12, ab = 32
প্রদত্ত রাশি,
2{a^2} + 2{b^2}
= 2({a^2} + {b^2})
= 2\{ {(a + b)^2} - 2ab\}
= 2\{ {(12)^2} - 2 \times 32\}
= 2(144 - 64)
= 2 \times 80
= 160 
2{a^2} + 2{b^2} = 160    (Ans.)

সমাধান (গ):
প্রমান করতে হবে যে, {(a + 2b)^2} - 5{b^2} = 176
এখন,
{(a + 2b)^2} - 5{b^2}
= {a^2} + 2 \times a \times 2b + {(2b)^2} - 5{b^2}
= {a^2} + 4ab + 4{b^2} - 5{b^2}
= {a^2} + 4ab - {b^2}
= {a^2} - {b^2} + 4ab
= (a + b)(a - b) + 4ab
= 12 \times 4 + 4 \times 32
= 48 + 128
= 176

এখানে,
{(a - b)^2} = {(a + b)^2} - 4ab
.           = {(12)^2} - 4 \times 32
.           = 144 - 128
.           = 16
বা, a - b = \sqrt {16}
\therefore a - b = 4

\therefore {(a + 2b)^2} - 5{b^2} = 176   (প্রমাণিত)


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