Question

How do you find two positive numbers such that the sum of the squares of the two numbers is 169 and the difference between the two numbers is 7?

Answer 1

Hint: To find two positive numbers, we need to consider x and y as the two unknown numbers, in which we need to form equations based on the given data and then solve those equations, in which we obtain the Quadratic equation of the form \[a{x^2} + bx + c\] , hence then we have to find the factors of the equation to get the value of the two numbers.
Formula used:
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]

Complete step by step solution:
Sum of the squares of the two numbers is 169 i.e., and the difference between the two numbers is 7.
Let x and y be the two positive numbers such that x be the larger of the 2 positive numbers and let y be the smaller of the 2 positive numbers.
As, given sum of the squares of the two numbers is 169 i.e.,
\[{x^2} + {y^2} = 169\] …………………… 1
The difference between the two numbers is 7 i.e.,
\[x - y = 7\] ………………………… 2
\[ \Rightarrow x = y + 7\] ……………………… 3
Now, substitute the value of x from equation 3 in equation 1 as:
\[{x^2} + {y^2} = 169\]
\[ \Rightarrow {\left( {y + 7} \right)^2} + {y^2} = 169\] …………………. 4
As, we obtained the equation of the form,\[{\left( {a + b} \right)^2}\] , in which expanding this we get \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] , hence apply this in equation 4 as:
\[ \Rightarrow {y^2} + 2\left( y \right)\left( 7 \right) + {\left( 7 \right)^2} + {y^2} = 169\]
Evaluating the terms, we get:
\[ \Rightarrow {y^2} + 14y + 49 + {y^2} = 169\]
Simplify and combine all the like terms, as:
\[ \Rightarrow 2{y^2} + 14y + 49 - 169 = 0\]
\[ \Rightarrow 2{y^2} + 14y - 120 = 0\]
As, 2 is common term in the obtained equation, so combining the terms we get:
\[ \Rightarrow 2\left( {{y^2} + 7y - 60} \right) = 0\]
Now, we need to find the factors as the obtained quadratic equation of the form \[a{x^2} + bx + c\] , by which we can easily find the factors of the equation using the AC method.
\[{y^2} + 7y - 60 = 0\]
The pair of integers we need to find for the product is c and whose sum is b, in which the product is -60 and sum is 7.
\[ \Rightarrow {y^2} + 12y - 5y - 60 = 0\]
Hence, the factors are:
\[ \Rightarrow \left( {y + 12} \right)\left( {y - 5} \right) = 0\] ………………… 5
Now, set individual factor of the equation 5 to zero and solve for it as:
\[\left( {y + 12} \right) = 0\]
\[\left( {y - 5} \right) = 0\]
Now let us solve for the first factor i.e.,
\[\left( {y + 12} \right) = 0\]
We, get:
\[y = - 12\]
Now let us solve for the second factor i.e.,
\[\left( {y - 5} \right) = 0\]
We, get:
\[y = 5\]
Hence, we got the values of y as, \[\left( {y = - 12,5} \right)\] however,\[y > 0\] , hence let us take the value of y as 5.
Now, substitute the value of y in equation 2, as:
\[x - y = 7\]
\[ \Rightarrow x - 5 = 7\]
\[ \Rightarrow x = 7 + 5 = 12\]
Hence, the value of x is 12.
Therefore, the two positive numbers are 12 and 5 i.e., \[\left( {x = 12,y = 5} \right)\] .
So, the correct answer is “\[\left( {x = 12,y = 5} \right)\]”.

Note: The key point to find two positive numbers is that we need to form the equations properly based on the given data and solving the equations, we get the obtained equation of the form \[a{x^2} + bx + c\] ,i.e., you must know what type of the equation it is; for example: Quadratic equation, Simultaneous Linear equation etc. And here, the obtained equation is a Quadratic equation.