Question

solve x^2+(X+1)^2=41​

Answer 1

x^2+x^2+1+2x=41
2x^2+2x-40=0
divide by 2 in whole eq.
x^2+x-20=0
x^2-4x+5x-20=0
x(x-4) + 5(x-4)
(x+5) (x-4)
[x=4 and -5]

Answer 2

$<b >HEYA !!!$

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x^2 + ( x + 1 )^2 = 41

x^2 + x^2 + 1 + 2x = 41

2x^2 + 2x = 41 - 1

2x^2 + 2x = 40

2( x^2 + x) = 20

x^2 + x -20 = 0

x^2 + 5x - 4x - 20 = 0

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

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

So ,

x = 4 & -5 ✔

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