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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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