Question

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

Answer 1

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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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Answer 2

${x}^{2} + {( x + 1) }^{2} =365</p><p>$

$↝ {x}^{2} + {x}^{2} + 2x + 1 = 365$

$↝2 {x}^{2} + 2x - 364 = 0$

$↝ {x}^{2} + x - 182 = 0$

$↝ {x}^{2} + 14x - 13x - 182 = 0$

$↝x(x + 14) - 13(x + 14) = 0$

$↝(x + 14)(x - 13) = 0$

$↝x + 14 = 0$

$↝x - 13 = 0$

$↝x = - 14$

$↝x = 13.$

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