Question

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

Answer:

QUESTION:−

Q:-factorise and hence find roots

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

2

+(x+1)

2

=365

$\huge\tt\underline\blue{Answer } </p><p>$

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

(x + 14)(x - 13) = 0 ; x_{1}x

1

= - 14 , x_{2}x

2

= 13

Step-by-step explanation:

x² + x² + 2x + 1 - 365 = 0

2x² + 2x - 364 = 0

x² + (1) x - 182 = 0

14 + (- 13) = 1 and 14 × (- 13) = - 182 ⇒ (x + 14)(x - 13) = 0

x_{1}x

1

= - 14

x_{2}x

2

= 13

Answer 2

Answer:

${x}^{2} + (x + 1 {)}^{2} = 365 \\ {x}^{2} + {x}^{2} + 2x + 1 = 365 \\ 2 {x}^{2} + 2x - 364 = 0 \\ 2( {x}^{2} + x - 182) = 0 \\ {x}^{2} + x - 182 = 0 \\ {x}^{2} + 14x - 13x - 182 \\ x(x + 14) - 13(x + 14) \\ (x + 14)(x - 13) \\ roots \: are \\ x = - 14 \\ x = 13$