Question

find the two consecutive positive integer s who sum of their square 365​

Answer 1

Solution: Let first number be x

Let second number be (x+1)

According to given condition, we have

x2+(x+1)2=365 {(a+b)2=a2+b2+2ab}

⇒x2+x2+1+2x=365

⇒2x2+2x−364=0

Dividing equation by 2, we get

x2+x−182=0

⇒x2+14x−13x−182=0

=> x(x+14) -13(x+14) = 0

=> (x-13)(x+14) =0

=> x= -14,13

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