Question

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

Answer 1

⟹x

2

+(x+1)

2

=365

⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365⟹x

2

+x

2

+2x+1=365

⟹2 {x}^{2} + 2x - 364 = 0⟹2x

2

+2x−364=0

⟹ {x}^{2} + x - 182 = 0⟹x

2

+x−182=0

⟹ {x}^{2} + 14x - 13x - 182 = 0⟹x

2

+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 ✓

Answer 2

Answer:

⟹x

2

+(x+1)

2

=365

⟹ {x}^{2} + {x}^{2} + 2x + 1 = 365⟹x

2

+x

2

+2x+1=365

⟹2 {x}^{2} + 2x - 364 = 0⟹2x

2

+2x−364=0

⟹ {x}^{2} + x - 182 = 0⟹x

2

+x−182=0

⟹ {x}^{2} + 14x - 13x - 182 = 0⟹x

2

+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 ✓