Question

Please solve 19th and 20th Question

Answer 1

Answer:

20) i) (2a+b)³

     ii) (4m-7n)(16m²+49n²+28mn)

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Answer 2

$\rule{200}{2}$

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$\huge\sf\underline\red{19^{th}\: Question\::}$

Find the value of a and b if

$\sf\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } + \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + b \sqrt{5}$

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$\huge\sf\underline\green{ Solution\::}$

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$\longrightarrow\sf( \frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } \times \frac{3 - \sqrt{5} }{3 - \sqrt{5} } ) - ( \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } ) = a + b \sqrt{5}$

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$\longrightarrow\sf( \frac{21 - 7 \sqrt{5 } + 9 \sqrt{5} - 15}{9 - 5} ) - ( \frac{21 + 7 \sqrt{5} - 9 \sqrt{5} - 15 }{9 - 5} )$

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$\longrightarrow\sf\frac{\cancel{21} - 7 \sqrt{5} + 9 \sqrt{5} - \cancel{15} - \cancel{21} - 7 \sqrt{5} + 9 \sqrt{5}+\cancel{15} }{4} = a + b \sqrt{5}$

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$\longrightarrow\sf\frac{18 \sqrt{5} - 14 \sqrt{5} }{4}$

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$\longrightarrow\sf\frac{4 \sqrt{5} }{4}$

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$\longrightarrow\sf\sqrt{5} = a + b \sqrt{5}$

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• $\large\sf {a\:=\:0}$
• $\large\sf {b\:=\:1}$

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$\rule{200}{2}$

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$\huge\sf\underline\red{20^{th}\: Question\::}$

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Factorise :

• 8a³+b³+12a²b+6ab²
• 64m³-343n³

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$\huge\sf\underline\green{ Solution\::}$

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① 8a³ + b³ + 12a²b + 6ab²

⇒ (2a+b) × (4a²-2ab-b²) + 6ab × (2a+b)

⇒ (2a + b) × (4a² - 2ab - b² + 6ab)

⇒ (2a + b) × (4a² + 4ab + b²)

⇒ (2a + b) × (2a + b)²

⇒ (2a + b)²

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② 64m³ - 343n³

⇒ 4³m³ - 7³n³

⇒ (4m)³ - (7n)³

⇒ (4m - 7n) × [(4m)² + 4m × 7n + (7n)²]

⇒ (4m - 7n) × (16m² + 28mn + 49n²)

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$\rule{200}{2}$