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Divide 29 into two parts so that the sum of the squares of the parts is 425.
Given:
The sum of the squares of the two parts of 29 is 425.
To do:
We have to divide 29 into two parts so that the sum of the squares of the parts is 425.
Solution:
Let one of the parts be $x$.
This implies, the other part is $29-x$.
According to the question,
$x^2+(29-x)^2=425$
$x^2+(29)^2-2(29)(x)+x^2=425$ (Since $(a-b)^2=a^2-2ab+b^2$)
$2x^2-58x+841=425$
$2x^2-58x+841-425=0$
$2x^2-58x+416=0$
$2(x^2-29x+208)=0$
$x^2-29x+208=0$
Solving for $x$ by factorization method, we get,
$x^2-13x-16x+208=0$
$x(x-13)-16(x-13)=0$
$(x-13)(x-16)=0$
$x-13=0$ or $x-16=0$
$x=13$ or $x=16$
If $x=13$, $29-x=29-13=16$
Therefore, the two parts are $13$ and $16$.
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