Divide 29 into two parts so that the sum of the square of the parts is 425.
Last updated date: 16th Mar 2023
•
Total views: 306k
•
Views today: 3.85k
Answer
306k+ views
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
