Question

The sum of the squares of two consecutive positive integers is 13. How do you find the integers?

Answer 1

Hint: Now let the two consecutive numbers be x and x + 1. Now we are given that the sum of squares of the numbers is 13. Hence we will use the condition to form a quadratic equation and simplify the equation in the form $a{{x}^{2}}+bx+c=0$. Now we know that the roots of the equation are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we get the value of x for which the condition is true. Hence we have the two consecutive numbers.

Complete step by step solution:
For the given question we are given to find two integers and it is given that the sum of the squares of two consecutive integers is 13.
As we know consecutive numbers means numbers which come in row.
Let us consider the two numbers as \[x\text{ , x+1}\].
Therefore equation from the question will be
\[ {{\left( x \right)}^{2}}+{{\left( x+1 \right)}^{2}}=13\]
Let us consider the above equation as equation (1).
\[\Rightarrow {{\left( x \right)}^{2}}+{{\left( x+1 \right)}^{2}}=13........\left( 1 \right)\]
Now let us expand the equation (1) by using the formula\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
Let us apply the formula to the equation (1), we get
\[\Rightarrow {{x}^{2}}+{{x}^{2}}+1+2x=13\]
Let us consider the equation as equation (2).
\[\Rightarrow {{x}^{2}}+{{x}^{2}}+1+2x=13.................\left( 2 \right)\]
By simplifying the equation (2), we get
\[\Rightarrow 2{{x}^{2}}+1+2x=13\]
Let us consider the above equation (3)
\[\Rightarrow 2{{x}^{2}}+1+2x=13............\left( 3 \right)\]
Subtract with 13 on both sides of the equation (3), we get
\[\Rightarrow 2{{x}^{2}}+1+2x-13=13-13\]
\[\Rightarrow 2{{x}^{2}}+2x-12=0\]
Let us consider the above equation as equation (4), we get
\[\Rightarrow 2{{x}^{2}}+2x-12=0...............\left( 4 \right)\]
Let us take ‘2’ as common from the equation (4), we get
\[\Rightarrow 2\left( {{x}^{2}}+x-6 \right)=0\]
Dividing the above equation with ‘2’, we get
\[\Rightarrow {{x}^{2}}+x-6=0\]
Let us consider the above equation as equation (5), we get
\[\Rightarrow {{x}^{2}}+x-6..............\left( 5 \right)\]
By applying the formula \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to the equation (5), we get
\[\begin{align}
  & \Rightarrow \dfrac{-1\pm \sqrt{1-4\left( 1 \right)\left( -6 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow \dfrac{-1\pm \sqrt{1+24}}{2\left( 1 \right)} \\
 & \Rightarrow \dfrac{-1\pm \sqrt{25}}{2\left( 1 \right)} \\
 & \Rightarrow \dfrac{-1\pm 5}{2} \\
 & \Rightarrow \dfrac{4}{2}\text{ or } \dfrac{-6}{2} \\
 & \Rightarrow 2\text{ or } \text{-3} \\
\end{align}\]
Therefore the values of x are 2 and -3, but in the given question it is given that they are positive integers. So the value of \[x=2\].

Therefore two consecutive numbers will be \[2\text{ and 3}\].

Note: Now note that we can find the roots of the quadratic equation by splitting the middle term also. Here we will split the middle term such that the product of the two terms is the product of the first term and the last term. Hence writing the middle term of the equation which is x as 3x – 2x we can simplify the equation and solve the equation.