Question

The square root of $71\times 72\times 73\times 74+1$ is ___.
A. 9375
B. 9625
C. 5625
D. 5255

Answer 1

Hint: To solve the square root of the given number, first we will observe the equation. by observing the equation we can find that the given numbers are the 4 consecutive numbers +1. i.e. $x\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)+1$ and simplify the equation. By assuming $x=71$ we can easily find the square root of a given number.

Complete step-by-step solution -
Here, in this question we asked to find the square root of $71\times 72\times 73\times 74+1$.
By observing this we can write it as $71\times \left( 71+1 \right)\times \left( 71+2 \right)\times \left( 71+3 \right)+1$
Now, we can consider it as the product of 4 consecutive numbers +1.
So, let us take $x=71$
Then the 4 consecutive number +1 be $x\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)+1$
Now, we will multiply the first term with last term and second term with third term, so we get –
$\begin{align}
  & \Rightarrow \left( x\left( x+3 \right) \right)\left( \left( x+1 \right)\left( x+2 \right) \right)+1 \\
 & =\left( {{x}^{2}}+3x \right)\left( {{x}^{2}}+2x+x+2 \right)+1 \\
 & =\left( {{x}^{2}}+3x \right)\left( {{x}^{2}}+3x+2 \right)+1 \\
\end{align}$
By multiplying the each terms of second term with first term, we get –
$\Rightarrow \left[ \left( {{x}^{2}}+3x \right){{x}^{2}}+\left( {{x}^{2}}+3x \right)3x+\left( {{x}^{2}}+3x \right)2 \right]+1$
By taking the product of common factors, we get –
$\begin{align}
  & \Rightarrow \left( \left( {{x}^{2}}+3x \right)\left( {{x}^{2}}+3x \right)+\left( {{x}^{2}}+3x \right)2 \right)+1 \\
 & =\left( {{\left( {{x}^{2}}+3x \right)}^{2}}+2\left( {{x}^{2}}+3x \right)+1 \right) \\
\end{align}$
We know that ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$
By using this identity, we get –
$\Rightarrow {{\left( \left( {{x}^{2}}+3x \right)+1 \right)}^{2}}$
$\therefore 71\times 72\times 73\times 74+1={{\left( \left( {{x}^{2}}+3x \right)+1 \right)}^{2}}$
Now, we will find the square root of \[71\times 72\times 73\times 74+1\]
$\Rightarrow \sqrt{71\times 72\times 73\times 74+1}=\sqrt{{{\left( \left( {{x}^{2}}+3x \right)+1 \right)}^{2}}}$
By cancelling the square root and whole square, we get –
$={{x}^{2}}+3x+1$
By substituting $x=71$ in the above equation, we get –
$\begin{align}
  & \Rightarrow {{\left( 71 \right)}^{2}}+3\left( 71 \right)+1 \\
 & =5041+213+1 \\
 & =5255 \\
\end{align}$
Hence, option D. is the correct answer.

Note: Students can solve this question in a simple way. For this they must know the perfect square of 3 and 4 digit numbers.
$\begin{align}
  & \Rightarrow \sqrt{71\times 72\times 73\times 74+1} \\
 & =\sqrt{27615025} \\
\end{align}$
To find the square root of 27615025 students must know the perfect root 5255.
Where ,
$\begin{align}
  & {{5255}^{2}}=5255\times 5255 \\
 & =27615025 \\
\end{align}$
So the $\sqrt{27615025}=\sqrt{5255\times 5255}=5255$ .