Question

Solve in positive integers:
$5{x^2} - 10xy + 7{y^2} = 77$

Answer 1

Hint: In this particular question use the concept of hit and trial method by substituting y = 1, 2, 3.... in the given equation until we get the smallest positive integer for x so use this concept to reach the solution of the question.

Complete step-by-step solution:
Given equation
$5{x^2} - 10xy + 7{y^2} = 77$.................. (1)
Now we have to find the solution of the above equation in positive integers.
So the simplest method to solve this is by hit and trial method.
As we need positive integers, so we cannot substitute y = 0.
So start substituting, y = 1, 2, 3, ..... In equation (1) until we get the positive integer for x.
So when, y = 1
So from equation (1) we have,
$ \Rightarrow 5{x^2} - 10x\left( 1 \right) + 7{\left( 1 \right)^2} = 77$
Now simplify it we have,
$ \Rightarrow 5{x^2} - 10x - 70 = 0$
Now divide by 5 throughout we have,
$ \Rightarrow {x^2} - 2x - 14 = 0$
So this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 1, b = -2, c = -14
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 14} \right)} }}{{2\left( 1 \right)}} = \dfrac{{2 \pm \sqrt {60} }}{2} = 1 \pm \sqrt {15} $
So square root of 15 is not an integer value so it is not the required solution.
Now when, y = 2
So from equation (1) we have,
$ \Rightarrow 5{x^2} - 10x\left( 2 \right) + 7{\left( 2 \right)^2} = 77$
Now simplify it we have,
$ \Rightarrow 5{x^2} - 20x - 49 = 0$
So this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 5, b = -20, c = -49
$ \Rightarrow x = \dfrac{{20 \pm \sqrt {{{\left( { - 20} \right)}^2} - 4\left( 5 \right)\left( { - 49} \right)} }}{{2\left( 5 \right)}} = \dfrac{{20 \pm \sqrt {1380} }}{{10}} = \dfrac{1}{5}\left( {10 \pm \sqrt {345} } \right)$
So square root of 345 is not an integer value so it is not the required solution.
Now when, y = 3
So from equation (1) we have,
$ \Rightarrow 5{x^2} - 10x\left( 3 \right) + 7{\left( 3 \right)^2} = 77$
Now simplify it we have,
$ \Rightarrow 5{x^2} - 30x - 14 = 0$
So this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 5, b = -30, c = -14
$ \Rightarrow x = \dfrac{{30 \pm \sqrt {{{\left( { - 30} \right)}^2} - 4\left( 5 \right)\left( { - 14} \right)} }}{{2\left( 5 \right)}} = \dfrac{{30 \pm \sqrt {1180} }}{{10}} = \dfrac{1}{5}\left( {15 \pm \sqrt {295} } \right)$
So square root of 295 is not an integer value so it is not the required solution.
Now when, y = 4
So from equation (1) we have,
$ \Rightarrow 5{x^2} - 10x\left( 4 \right) + 7{\left( 4 \right)^2} = 77$
Now simplify it we have,
$ \Rightarrow 5{x^2} - 40x + 35 = 0$
So this is a quadratic equation so apply quadratic formula we have,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 5, b = -40, c = 35
$ \Rightarrow x = \dfrac{{40 \pm \sqrt {{{\left( { - 30} \right)}^2} - 4\left( 5 \right)\left( {35} \right)} }}{{2\left( 5 \right)}} = \dfrac{{40 \pm \sqrt {900} }}{{10}} = \dfrac{{40 \pm 30}}{{10}} = 4 \pm 3$
So the possible values of x is (4 + 3), (4 - 3)
Therefore, x = 7, 1
So the solution of the given equation in positive integers are (1, 4) and (7, 4).

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that 0 is neither consider as positive nor negative, and the basis to solve these types of problem quickly is by hit and trial method, so substitute the values of y starting from 1 in the given equation and solve for x as above we will get the required solution in positive integers.