Question

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

Answer 1

x²+(x+1)²=61

x²+x²+2x+1=61

2x²+2x+1-61=0

2x²+2x-60=0

x²+x-30=0

(x+6)(x-5)=0

x=-6,5

x²=-6²=36

(x+1)²=7²=49

x²=5²=25

(x+1)²=(5+1)²=6²=36

hope it helps..

Answer 2

$\mathfrak{\huge{\blue{\underline{\underline{ANSWER :}}}}}$

We are given that the sum of the squares of two consecutive numbers is equal to 61. Let the numbers be x and x+1.

So x^2 + (x+1)^2 = 61

=> x^2 + x^2 + 1 + 2x = 61

=> 2x^2 + 2x - 60 = 0

=> x^2 + x - 30 =0

=> x^2 + 6x - 5x - 30 =0

=> x(x+6) -5(x+6) =0

=>(x-5)*(x+6) =0

We can have x = 5 and x = -6

Therefore the numbers can be 5 and 6 or -6 and -5.

The required result is (5, 6) and (-6,-5).

Hope It Will Help.❤️