Question

The sum of the square of a number and the square of its very next number is 421. Find those numbers.

Answer 1

Hint: We first assume a variable for the number and its next number. Then we use the given condition to find the mathematical expression. We solve the quadratic to find the solutions of the problem.

Complete step by step solution:
We try to assume a variable for the number as $x$.
The very next number is $x+1$.
It is given that the sum of the square of a number and the square of its very next number is 421.
The square of the number is ${{x}^{2}}$ and the square of the next number is ${{\left( x+1 \right)}^{2}}$.
The sum is 421.
Therefore, ${{x}^{2}}+{{\left( x+1 \right)}^{2}}=421$. This is the mathematical expression of the given condition.
Now we simplify the equation to get
$\begin{align}
  & {{x}^{2}}+{{\left( x+1 \right)}^{2}}=421 \\
 & \Rightarrow {{x}^{2}}+{{x}^{2}}+2x+1=421 \\
 & \Rightarrow 2{{x}^{2}}+2x-420=0 \\
 & \Rightarrow {{x}^{2}}+x-210=0 \\
\end{align}$
Now we have to factorise the quadratic to find its solution.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+x-210=0$. The values of a, b, c is $1,1,-210$ respectively.
We put the values and get x as $x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 1\times \left( -210 \right)}}{2\times 1}=\dfrac{-1\pm \sqrt{841}}{2}=\dfrac{-1\pm 29}{2}=14,-15$.
Now if we take $x=14$, the next number is $x=14+1=15$ and if we take $x=-15$, the next number is $x=-15+1=-14$.
The possible number duals can be $\left( 14,15 \right);\left( -15,-14 \right)$.

Note: For the quadratic we also could have broken it in the form of \[{{x}^{2}}+x-210={{x}^{2}}+15x-14x-210\].
We then use the grouping method to solve the equation where
\[\begin{align}
  & {{x}^{2}}+x-210 \\
 & ={{x}^{2}}+15x-14x-210 \\
 & =\left( x+15 \right)\left( x-14 \right) \\
\end{align}\]
Therefore, \[\left( x+15 \right)\left( x-14 \right)=0\] gives $x=14,-15$.