Question

Three consecutive positive integers are such that the sum of the square of the first and the product of other two is $46$, The smallest integer is:

Answer 1

$\Rightarrow$ Let there consecutive positive integers are $x,x+1,x+2.$

According to the question,

$\Rightarrow$ $x^2+(x+1)(x+2)=46$

$\Rightarrow$ $x^2+x^2+2x+x+2=46$

$\Rightarrow$ $2x^2+3x+2-46=0$

$\Rightarrow$ $2x^2+3x-44=0$

$\Rightarrow$ Here, $a=2,b=3$ and $c=44$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow$ $x=\frac{-3\pm \sqrt{(3{)}^2-4(2\left)\right(-44)}}{2\left(2\right)}$

$\Rightarrow$ $x=\frac{-3\pm \sqrt{361}}{4}$

$\Rightarrow$ $x=\frac{-3\pm 19}{4}$

$\Rightarrow$ $x=\frac{-3+19}{4}$ and $x=\frac{-3-19}{4}$

$\therefore$ $x=\frac{16}{4}$ and $x=\frac{-22}{4}$

$\therefore$ $x=4$ and $x=\frac{-11}{2}$

$\Rightarrow$ Here, there is only one integer which is $4$.

$\therefore$ Smallest integer is $4$.