# Divide 29 into two parts so that the sum of the square of the parts is 425.

Last updated date: 16th Mar 2023

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Hint: Let the first number be $x$ and the second number be $\left( {29 - x} \right)$, so that the sum of their squares is 425, so use this concept to get the solution of the question.

Given number is 29.

Now we have to divide 29 into two parts so that the sum of the square of the parts is 425.

So, let the first number be $x$ and second number be $\left( {29 - x} \right)$

Now according to given condition

${x^2} + {\left( {x - 29} \right)^2} = 425$

Now we know ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so, apply this property in above equation we have

$

{x^2} + {x^2} + {29^2} - 2 \times 29 \times x = 425 \\

\Rightarrow 2{x^2} - 58x + 841 - 425 = 0 \\

\Rightarrow 2{x^2} - 58x + 416 = 0 \\

$

Now divide by 2 in above equation we have

${x^2} - 29x + 208 = 0$

Now factorize the above equation

$

\Rightarrow {x^2} - 16x - 13x + 208 = 0 \\

\Rightarrow x\left( {x - 16} \right) - 13\left( {x - 16} \right) = 0 \\

\Rightarrow \left( {x - 16} \right)\left( {x - 13} \right) = 0 \\

\therefore \left( {x - 16} \right) = 0{\text{ & }}\,{\text{ }}\left( {x - 13} \right) = 0 \\

\therefore x = 16{\text{ & }}x = 13 \\

$

So, these are the required numbers which divide 29 into two parts so that the sum of the square of the parts is 425.

Note: In such types of question the key concept is that always assume the first number be $x$ and second number be $\left( {{\text{value}} - x} \right)$, then according to given condition simplify, we will get the required numbers which divide 29 into two parts so that the sum of the square of the parts is 425.

Given number is 29.

Now we have to divide 29 into two parts so that the sum of the square of the parts is 425.

So, let the first number be $x$ and second number be $\left( {29 - x} \right)$

Now according to given condition

${x^2} + {\left( {x - 29} \right)^2} = 425$

Now we know ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so, apply this property in above equation we have

$

{x^2} + {x^2} + {29^2} - 2 \times 29 \times x = 425 \\

\Rightarrow 2{x^2} - 58x + 841 - 425 = 0 \\

\Rightarrow 2{x^2} - 58x + 416 = 0 \\

$

Now divide by 2 in above equation we have

${x^2} - 29x + 208 = 0$

Now factorize the above equation

$

\Rightarrow {x^2} - 16x - 13x + 208 = 0 \\

\Rightarrow x\left( {x - 16} \right) - 13\left( {x - 16} \right) = 0 \\

\Rightarrow \left( {x - 16} \right)\left( {x - 13} \right) = 0 \\

\therefore \left( {x - 16} \right) = 0{\text{ & }}\,{\text{ }}\left( {x - 13} \right) = 0 \\

\therefore x = 16{\text{ & }}x = 13 \\

$

So, these are the required numbers which divide 29 into two parts so that the sum of the square of the parts is 425.

Note: In such types of question the key concept is that always assume the first number be $x$ and second number be $\left( {{\text{value}} - x} \right)$, then according to given condition simplify, we will get the required numbers which divide 29 into two parts so that the sum of the square of the parts is 425.

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