Question

Q:-factorise and hence find roots
${x}^{2} + {(x + 1) }^{2} = 365$
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Answer 1

Step-by-step explanation:

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-factorise and hence find roots

${x}^{2} + {(x + 1) }^{2} = 365$

$\huge\tt\underline\blue{Answer }$

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$⟹ {x}^{2} + {(x + 1)}^{2} = 365$

$⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365$

$⟹2 {x}^{2} + 2x - 364 = 0$

$⟹ {x}^{2} + x - 182 = 0$

$⟹ {x}^{2} + 14x - 13x - 182 = 0$

$⟹x(x + 14) - 13(x + 14) = 0$

$⟹(x + 14)(x - 13) = 0$

$⟹x + 14 = 0$

$⟹x - 13 = 0$

$⟹x = - 14$✓

$⟹x = 13$✓

HOPE IT HELPS YOU..

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Thankyou:)

Answer 2

$\sf \huge \blue {Answer}$

$\sf {x}^{2} + {(x + 1)}^{2} = 365$

$\sf {x}^{2} + {x}^{2} + 2x + 1 = 365$

$\sf 2 {x}^{2} + 2x - 364 = 0$

$\sf {x}^{2} + x - 182 = 0$

$\sf {x}^{2} + 14x - 13x - 182 = 0$

$\sf x(x + 14) - 13(x + 14) = 0$

$\sf (x + 14)(x - 13) = 0$

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$\sf x + 14 = 0$

$\boxed{\sf x = - 14}$

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$\sf x - 13 = 0$

$\boxed {\sf x = 13}$

Therefore, roots of x are -14 and 13