Question:
Divide 29 into two parts so that the sum of the squares of the parts is 425.
Solution:
Let first numbers be $x$ and other $(29-x)$
Then according to question
$x^{2}+(29-x)^{2}=425$
$x^{2}+x^{2}-58 x+841=425$
$2 x^{2}-58 x+841=425$
$2 x^{2}-58 x+841-425=0$
$2 x^{2}-58 x+416=0$
$x^{2}-29 x+208=0$
$x^{2}-16 x-13 x+208=0$
$x(x-16)-13(x-16)=0$
$(x-16)(x+13)=0$
$(x-16)=0$
$x=16$
Or
$(x+13)=0$
$x=-13$
Since, 29 being a positive number, so x cannot be negative.
Therefore,
When $x=16$ then
$29-x=29-16$
$=13$
Thus, two consecutive number be