Question

26)
Find two consecutive positive integers, sum of whose squares is41
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Answer 1

Heya Mate ur answer us..

Let the two nos. be x,x+1

=>x²+(x+1)²=41

=> 2x²+2x+1=41

=> 2x²+2x-40=0

=>x²+x-20=0

=>x²+5x-4x-20=0

(x+5)(x-4)=0

x=-5/4

So, nos are-----> -5,-4

4,5

Hope it helps ❤❤❤❤

Answer 2

Let the first integer = x

Let the second integer = x + 1

According to question :

(x)² + (x + 1)² = 41

x² + x² + 2x + 1 = 41

2x² + 2x = 41 - 1

2x² + 2x = 40

2(x² + x) = 40

x² + x = 40/2

x² + x = 20

x² + x - 20 = 0

x² + (5 - 4)x - 20 = 0

x² + 5x - 4x - 20 = 0

x(x + 5) -4(x + 5) = 0

(x - 4) (x + 5) = 0

x = 4 or x = (-5)

It is given that the integer is positive so :

The first integer = 4

The second integer = 5

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