Question

The sum of two numbers is 8 and the sum of their squares is 261. How do you find the numbers?

Answer 1

Hint: In this question we have to find the numbers according to the values given, for this first assume the two numbers be \[x\] and \[y\] , and sum of squares can be written as \[{x^2} + {y^2}\] , and using the algebraic identities \[{\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\] and \[{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2}\] and simplifying the given equations we will get the required result.

Complete step by step solution:
Given that the sum of two numbers is 8 and the sum of their squares is 261.
Let us consider the two numbers be \[x\] and \[y\] ,
So, from the given data we get,
\[ \Rightarrow \] \[x + y = 9 - - - - (1)\] , and
\[{x^2} + {y^2} = 261\] ,
Now we know that\[{\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\] ,
By substitute the given values in the identity, we get,
\[ \Rightarrow {\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\] ,
Here \[x + y = 9\] and\[{x^2} + {y^2} = 261\] , substituting the values we get,
\[ \Rightarrow {9^2} = 261 + 2xy\] ,
Now simplifying we get,
\[ \Rightarrow 81 = 261 + 2xy\] ,
Now subtracting both sides with 261 we get,
\[ \Rightarrow 81 - 261 = 261 + 2xy - 261\] ,
Now simplifying we get,
\[ \Rightarrow 2xy = - 180\] ,
Now dividing both sides with 2 we get,
\[ \Rightarrow \dfrac{{2xy}}{2} = \dfrac{{ - 180}}{2}\] ,
Now simplifying we get,
\[ \Rightarrow xy = - 90\] ,
Now using the identity, \[{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2}\] , and substituting the values in the formula , we get,
\[ \Rightarrow {\left( {x - y} \right)^2} = 261 - 2\left( { - 90} \right)\] ,
Now simplifying we get,
\[ \Rightarrow {\left( {x - y} \right)^2} = 261 + 180\] ,
Now adding on the right hand side we get,
\[ \Rightarrow {\left( {x - y} \right)^2} = 441\] ,
Now taking out the square, we get,
\[ \Rightarrow \left( {x - y} \right) = \sqrt {441} \] ,
Now applying the square root we get,
\[ \Rightarrow x - y = \pm 21 - - - - (2)\] ,
Now solving equations (1) and (2) first take \[x - y = 21\] we get,
\[ \Rightarrow 2x = 30\] ,
Now divide both sides with 2 we get,
\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{30}}{2}\] ,
Now simplifying we get,
\[ \Rightarrow x = 15\] ,
Now substituting the value of \[x\] in (1) we get,
\[ \Rightarrow 15 + y = 9\] ,
Now subtract both sides with 15 we get,
\[ \Rightarrow 15 + y - 15 = 9 - 15\] ,
Now simplifying we get,
\[ \Rightarrow y = - 6\] ,
So, the numbers are \[x = 15\] and \[y = - 6\] ,
Now solving equations (1) and (2) first take \[x - y = - 21\] we get,
\[ \Rightarrow 2x = - 12\] ,
Now divide both sides with 2 we get,
\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{ - 12}}{2}\] ,
Now simplifying we get,
\[ \Rightarrow x = - 6\] ,
Now substituting the value of \[x\] in (1) we get,
\[ \Rightarrow - 6 + y = 9\] ,
Now adding both sides with 6 we get,
\[ \Rightarrow - 6 + y + 6 = 9 + 6\] ,
Now simplifying we get,
\[ \Rightarrow y = 15\] ,
So, the numbers are \[x = - 6\] and \[y = 15\] ,
So, by any means the numbers are 15 and -6.

The numbers whose sum of two numbers is 8 and the sum of their squares is 261 are equal to 15 and -6.

Note: Algebraic identities are algebraic equations which are always true for every value of variables in them. In an algebraic identity, the left-side of the equation is equal to the right-side of the equation, some of the algebraic identities that are commonly used are:
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] ,
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] ,
 \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] ,
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] ,
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\] .