Question

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

$⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365$

$⟹2 {x}^{2} + 2x - 364 = 0$

$⟹ {x}^{2} + x - 182 = 0$

$⟹ {x}^{2} + 14x - 13x - 182 = 0$

$⟹x(x + 14) - 13(x + 14) = 0$

$⟹(x + 14)(x - 13) = 0$

$⟹x + 14 = 0$

$⟹x - 13 = 0$

$⟹x = - 14$✓

$⟹x = 13$✓

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Answer 2

Answer:

365

Step-by-step explanation:

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⟹ {x}^{2} + {(x + 1)}^{2} = 365⟹x2+(x+1)2=365

⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365⟹x2+x2+2x+1=365

⟹2 {x}^{2} + 2x - 364 = 0⟹2x2+2x−364=0

⟹ {x}^{2} + x - 182 = 0⟹x2+x−182=0

⟹ {x}^{2} + 14x - 13x - 182 = 0⟹x2+14x−13x−182=0

⟹x(x + 14) - 13(x + 14) = 0⟹x(x+14)−13(x+14)=0

⟹(x + 14)(x - 13) = 0⟹(x+14)(x−13)=0

⟹x + 14 = 0⟹x+14=0

⟹x - 13 = 0⟹x−13=0

⟹x = - 14⟹x=−14 ✓

⟹x = 13⟹x=13 ✓

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