Find two consecutive natural numbers whose squares have the sum 265.
[a] -11 , -12
[b] 11, 12
[c] 10, 11
[d] -10,-9

Hint: Two consecutive numbers differ by 1. Hence if one of them is n, the other one is n+1. So, let the numbers be n,n+1. Form a quadratic in n and solve for n. Hence find the numbers. Check for extraneous roots. Verify your solution.

Complete step-by-step answer:

Let the numbers be n and n+1.
Now the sum of squares of the numbers $={{n}^{2}}+{{\left( n+1 \right)}^{2}}$
Using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get
The sum of the squares of the numbers
$\begin{align}
  & ={{n}^{2}}+{{n}^{2}}+2n+1 \\
 & =2{{n}^{2}}+2n+1 \\
\end{align}$
Given that the sum of the squares is 265.
Hence, we have
$2{{n}^{2}}+2n+1=265$
Subtracting 265 from both sides, we get
$\begin{align}
  & 2{{n}^{2}}+2n+1-265=265-265 \\
 & \Rightarrow 2{{n}^{2}}+2n-264=0 \\
\end{align}$
Dividing both sides by 2, we get
$\begin{align}
  & \dfrac{2{{n}^{2}}+2n-264}{2}=\dfrac{0}{2} \\
 & \Rightarrow {{n}^{2}}+n-132=0 \\
\end{align}$
We solve using splitting the middle term method.
We have $11\times 12=132$ and $12-11=1$
Hence, we have
${{n}^{2}}+n-132={{n}^{2}}+12n-11n-132$
Taking n common from first two terms and 11 common from last two terms, we get
${{n}^{2}}+n-132=n\left( n+12 \right)-11\left( n+12 \right)$
Taking n+12 common from both the terms, we get
${{n}^{2}}+n-132=\left( n+12 \right)\left( n-11 \right)$
Hence we have
$\begin{align}
  & {{n}^{2}}+n-132=0 \\
 & \Rightarrow \left( n+12 \right)\left( n-11 \right)=0 \\
\end{align}$
If ab = 0 then a =0 or b = 0 (zero product property).
Hence, we have
$\begin{align}
  & n+12=0\text{ or }n-11=0 \\
 & \Rightarrow n=-12\text{ or }n=11 \\
\end{align}$
Since n is a natural number, we have n>0.
Hence n =-12 is rejected
Hence n = 11.
Hence the numbers are 11 and 12.
Hence option [b] is correct.

Note: [1] Do not forget to delete the root n =-12 as it would give non-natural number solutions
[2] You can also use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which gives the roots of the quadratic $a{{x}^{2}}+bx+c=0$. Compare to get the value of a, b and c.
[3] Verification: $11\in \mathbb{N},12\in \mathbb{N},{{11}^{2}}+{{12}^{2}}=121+144=265$ and 11 and 12 are consecutive. Hence the solution is correct.