Find two consecutive natural nos the sum of whose squares is 145?
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let first no be x
second no is x+1
then
square of first no = x^2
square of second no =(x+1)^2 = x^2 +1 +2x
then
x^2+x^2+1+2x = 145
2x^2 +2x = 144
2x^2 +2x -144 =0
2 (x^2 +x-72) =0
2 (x^2 -8x+9x-72 =0
2 [x(x-8) -9(x-8)] =0
2 (x-9) (x-8) =0
now
these no are 8&9 (x-8=0 ,x=8)
second no is x+1
then
square of first no = x^2
square of second no =(x+1)^2 = x^2 +1 +2x
then
x^2+x^2+1+2x = 145
2x^2 +2x = 144
2x^2 +2x -144 =0
2 (x^2 +x-72) =0
2 (x^2 -8x+9x-72 =0
2 [x(x-8) -9(x-8)] =0
2 (x-9) (x-8) =0
now
these no are 8&9 (x-8=0 ,x=8)
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