Question

Find two consecutive integers such that the sum of their squares is 61.



Answer: 5,6 or -6,-5.



Plz give in detail solutions.

☺☺​

Answer 1

$answer :$

• Let the 2 integers be a , b

ATP ,

${a}^{2} + {b}^{2} = 61$

But ,

• Since , a and b are consecutive

$b = a + 1$

• By inserting this in eq(1) we get

${a}^{2} + {(a + 1)}^{2} = 61$

${a}^{2} + {a}^{2} + 1 + 2a = 61$

$2 {a}^{2} + 2a = 60$

${a}^{2} + a = 30$

${a}^{2} + a - 30 = 0$

${a}^{2} + 6a - 5a - 30 = 0$

$a(a + 6) - 5(a + 6) = 0$

$(a + 6)(a - 5) = 0$

$\implies \: a = - 6 \: (or )\: a = 5$

Therefore,

• The numbers are -6,-5 (or) 5,6