Question

2. Divide 29 into two parts so that the sum of the squares
of the parts is 425.​

Answer 1

Answer:

Let the two parts be x and y

x + y = 29 ⇒ y = 29 - x

so according to question x² + y² = 425

⇒ x² + (29 - x)² = 425

⇒ x² + 841 + x² - 58x = 425

⇒2x² - 58x + 416 = 0

⇒ x² - 29x + 208 = 0

⇒ x² - 13x - 16x + 208 = 0

⇒ x(x - 13) - 16(x - 13) = 0

⇒ (x - 16)(x -  13) = 0

 so x = 16 or 13  ANSWER

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• sum of 2 parts = 29
• sum of their squares = 425

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• find the numbers = ???

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

let the 2 parts be x and y

• sum of parts is 29

$\sf\implies x + y = 29$

$\sf\implies x = 29 - y$

• sum of the squares of the parts is 425.

$\sf\implies x^2 + y^2 = 425$

$\sf\implies ( 29 - y)^2 + y^2 = 425$

$\sf\implies 841 + y^2 - 58y + y^2 = 425$

$\sf\implies 841 + 2y^2 - 58y = 425$

$\sf\implies 841 - 425 + 2y^2 - 58y = 0$

$\sf\implies 416 + 2y^2 - 58y = 0$

$\sf\implies 2y^2 - 58y + 416 = 0$

$\sf\implies 2 ( y^2 - 29y + 208 ) = 0$

$\sf\implies y^2 - 29y + 208 = 0$

$\sf\implies y^2 - 13y - 16y + 208 = 0$

$\sf\implies y ( y - 13 ) - 16( y - 13) = 0$

$\sf\implies ( y - 16 ) ( y - 13 )= 0$

☆ $\sf y - 16 = 0$

$\sf\longrightarrow y = 16$

☆ $\sf y - 13 = 0$

$\sf\longrightarrow y = 13$

$\begin{tabular}{|c|c|}\cline{1-2}\sf if\: y \:= 16 &\sf if\: y \: = 13 \\\cline{1-2}\sf x + 16 = 29 &\sf x + 13 = 29 \\\cline{1-2}\sf x = 13 &\sf x = 16 \\\cline{1-2}\end{tabular}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Therefore 29 should be divided into 16 and 13 so that the sum of the squares of part is 425 .