Question

Find the three consecutive numbers such that the sum of their square will be $434$.

Answer 1

To determine the three consecutive numbers such that the sum of their square will be $434$:

Let the three consecutive numbers be $x,\left(x+1\right),\left(x+2\right)$.

Then the square of the numbers is $x^2,{\left(x+1\right)}^2,{\left(x+2\right)}^2$.

According to question, the sum of the square of the consecutive numbers is $434$. That is, $x^2+{\left(x+1\right)}^2+{\left(x+2\right)}^2=434$.

Solve the equation for $x$:

$\begin{array}{rcl}{x}^2+{\left(x+1\right)}^2+{\left(x+2\right)}^2& =& 434\\ x^2+x^2+1+2x+x^2+4+4x& =& 434\\ 3x^2+6x+5& =& 434\\ 3x^2+6x-429& =& 0\\ x^2+2x-143& =& 0\end{array}$

Solve the equation $x^2+2x-143=0$using splitting of middle term as follows:

$\begin{array}{rcl}{x}^2+2x-143& =& 0\\ x^2+13x-11x-143& =& 0\\ x\left(x+13\right)-11\left(x+13\right)& =& 0\\ \left(x+13\right)\left(x-11\right)& =& 0\end{array}$

If $\left(x+13\right)=0\Rightarrow x=-13$ and if $\left(x-11\right)=0\Rightarrow x=11$.

Hence, the value of $x$ is $11$or $-13$.

The required three consecutive numbers are:

$\begin{array}{rcl}x& =& 11\\ \left(x+1\right)& =& \left(11+1\right)=12\\ \left(x+2\right)& =& \left(11+2\right)=13\\ & & or\\ x& =& -13\\ \left(x+1\right)& =& \left(-13+1\right)=-12\\ \left(x+2\right)& =& \left(-13+2\right)=-11\end{array}$

Hence, the three consecutive numbers such that the sum of their square will be $434$ is $11,12,13\text{or}-13,-12,-11$.