Question

The sum of 2 numbers is 14 and the sum of their squares is 148. 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 = 14 - - - - (1)$, and
${x^2} + {y^2} = 148$,
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 = 14$and${x^2} + {y^2} = 148$, substituting the values we get,
$ \Rightarrow {14^2} = 148 + 2xy$,
Now simplifying we get,
$ \Rightarrow 196 = 148 + 2xy$,
Now subtracting both sides with 148 we get,
$ \Rightarrow 196 - 148 = 148 + 2xy - 148$,
Now simplifying we get,
$ \Rightarrow 2xy = 48$,
Now dividing both sides with 2 we get,
$ \Rightarrow \dfrac{{2xy}}{2} = \dfrac{{48}}{2}$,
Now simplifying we get,
$ \Rightarrow xy = 24$,
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} = 148 - 2\left( {24} \right)$,
Now simplifying we get,
$ \Rightarrow {\left( {x - y} \right)^2} = 148 - 48$,
Now adding on the right hand side we get,
$ \Rightarrow {\left( {x - y} \right)^2} = 100$,
Now taking out the square, we get,
$ \Rightarrow \left( {x - y} \right) = \sqrt {100} $,
Now applying the square root we get,
$ \Rightarrow x - y = \pm 10 - - - - (2)$,
Now solving equations (1) and (2) first take $x - y = 10$ we get,
$ \Rightarrow 2x = 24$,
Now divide both sides with 2 we get,
$ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{24}}{2}$,
Now simplifying we get,
$ \Rightarrow x = 12$,
Now substituting the value of $x$ in (1) we get,
$ \Rightarrow 12 + y = 14$,
Now subtract both sides with 12 we get,
$ \Rightarrow 12 + y - 12 = 14 - 12$,
Now simplifying we get,
$ \Rightarrow y = 2$,
So, the numbers are$x = 12$ and $y = 2$,
Now solving equations (1) and (2) first take $x - y = - 10$we get,
$ \Rightarrow 2x = 4$,
Now divide both sides with 2 we get,
$ \Rightarrow \dfrac{{2x}}{2} = \dfrac{4}{2}$,
Now simplifying we get,
$ \Rightarrow x = 2$,
Now substituting the value of$x$in (1) we get,
$ \Rightarrow 2 + y = 14$,
Now adding both sides with 2 we get,
$ \Rightarrow 2 + y - 2 = 14 - 2$
Now simplifying we get,
$ \Rightarrow y = 12$,
So, the numbers are $x = 2$ and $y = 12$

$\therefore $ The sum of 2 numbers is 14 and the sum of their squares is 148 are equal to 12 and 2.

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:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$,
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$,
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$,
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$,
${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$.