 QUESTION

# The value of $\sqrt {(28)(29)(30)(31) + 1}$ equals

Hint: Here to solve this question, we have to suppose $x = 29$, then $28 = x - 1,30 = x + 1,31 = x + 2$. This will help us in cancelling the square root, we are actually making it in the form of a perfect square to get an integer.

According to question,
$\Rightarrow \sqrt {(28)(29)(30)(31) + 1}$
$\Rightarrow \sqrt {(x - 1)(x)(x + 1)(x + 2) + 1}$
$\Rightarrow \sqrt {({x^2} + 2x)({x^2} - 1) + 1}$
$\Rightarrow \sqrt {{{({x^2})}^2} + 2(x)({x^2}) - {x^2} - 2x + 1}$, it can be rewritten as
$\Rightarrow \sqrt {{{({x^2})}^2} + {{(x)}^2} + {{( - 1)}^2} + 2({x^2})(x) + 2(x)( - 1) + 2( - 1)({x^2})}$ , now it is of the form $\sqrt {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca}$which is equal to $\sqrt {{{(a + b + c)}^2}}$where $a = {x^2}$, $b = x$ and $c = - 1$.
Now, under root will get cancelled and we are left with
$\Rightarrow \sqrt {{{({x^2} + x - 1)}^2}}$
$\Rightarrow {x^2} + x - 1$
$\Rightarrow x(x + 1) - 1$, now we will replace $x$ by 29 to get the required answer
We get, $\Rightarrow (29 \times 30) - 1$=869
Hence, the required result is 869.

Note: To form it in the form of ${(a + b + c)^2}$, which cancels the under root and make the calculation easier, we have to make ${x^2}$ as ${b^2}$so we have added $- 2{x^2}$. This question can also be done by direct multiplication method, but it will consume so much time, that is why we always use this method.