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Home » The sum of two numbers is     and the sum of their squares is     . Taking one number as x, form ail equation in x and solve it to find the numbers.
The sum of two numbers is

    \[9\]

and the sum of their squares is

    \[41\]

. Taking one number as x, form ail equation in x and solve it to find the numbers.
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The sum of two numbers is

    \[9\]

and the sum of their squares is

    \[41\]

. Taking one number as x, form ail equation in x and solve it to find the numbers.

Given:

Sum of two number =

    \[9\]

Let us consider first number be ‘x’

Second number be ‘

    \[9-x\]

’

So according to the question,

    \[\begin{array}{*{35}{l}} {{\left( x \right)}^{2}}~+\text{ }{{\left( 9\text{ }\text{ }x \right)}^{2}}~=\text{ }41 \\ {{x}^{2}}~+\text{ }81\text{ }\text{ }18x\text{ }+\text{ }{{x}^{2}}~\text{ }41\text{ }=\text{ }0 \\ 2{{x}^{2}}~\text{ }18x\text{ }+\text{ }40\text{ }=\text{ }0 \\ \end{array}\]

Divide by

    \[2\]

, we get

    \[{{x}^{2}}~\text{ }9x\text{ }+\text{ }20\text{ }=\text{ }0\]

Let us factorize,

    \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }4x\text{ }\text{ }5x\text{ }+\text{ }20\text{ }=\text{ }0 \\ x\left( x\text{ }\text{ }4 \right)\text{ }\text{ }5\left( x\text{ }\text{ }4 \right)\text{ }=\text{ }0 \\ \left( x\text{ }\text{ }4 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }0 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} \left( x\text{ }\text{ }4 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }0 \\ x\text{ }=\text{ }4\text{ }or\text{ }x\text{ }=\text{ }5 \\ \end{array}\]

When x =

    \[4\]

, then

First number = x =

    \[4\]

Second number =

    \[9\text{ }\text{ }x\text{ }=\text{ }9\text{ }\text{ }4\text{ }=\text{ }5\]

When x =

    \[5\]

, then

First number = x =

    \[5\]

Second number =

    \[9\text{ }\text{ }x\text{ }=\text{ }9\text{ }\text{ }5\text{ }=\text{ }4\]

∴ The required numbers are

    \[4\]

and

    \[5\]

.

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