The equation formed by decreasing each root of \[a{x^2} + bx + c = 0\] by \[1\] is \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\], then
A. \[\begin{array}{*{20}{c}}a& = &{ - b}\end{array}\]
B. \[\begin{array}{*{20}{c}}b& = &{ - c}\end{array}\]
C. \[\begin{array}{*{20}{c}}c& = &{ - a}\end{array}\]
D. \[\begin{array}{*{20}{c}}b& = &{a + c}\end{array}\]

Hint: In this question, first of all, we will determine the sum and the product of the roots of the equation\[\begin{array}{*{20}{c}}{a{x^2} + bx + c}& = &0\end{array}\]. After that, we will do the same for the equation \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\] and from this, we will get that the sum and the product of the roots of the equation are equal. And then we will satisfy the condition from the previous equation and hence, we will get a suitable answer.

Formula Used:
Root’s sum can be determined by
\[\alpha + \beta = \frac{{ - b}}{a}\]
Root’s product can be determined by
\[\alpha \beta = \frac{c}{a}\]

Complete step by step solution:
We have been provided in the question that,
The equation formed by decreasing each root of \[a{x^2} + bx + c = 0\] by \[1\] is
\[2{x^2} + 8x + 2 = 0\]
Let us assume that the roots of the above equation are \[\alpha \]and \[\beta \]respectively.
We know the sum and the product of the roots of the above equation.
Therefore, we can write
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\] -------- (1)
And
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha \beta }& = &{\frac{c}{a}}\end{array}\] ------------ (2)
 Now we have given that when the roots of the above equation are decreased by \[1\], then the equation \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\]is formed.
Therefore, the new roots will be \[\alpha - 1\]and \[\beta - 1\]respectively.
Now
The sum and the product of the roots of the formed equation will be
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right) + \left( {\beta - 1} \right)}& = &{ - \frac{8}{2}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - 2}\end{array}\]
And
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right)\left( {\beta - 1} \right)}& = &{\frac{2}{2}}\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta - \left( {\alpha + \beta } \right) + 1}& = &1\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta }& = &{\alpha + \beta }\end{array}\] -------- (3)
Now put the value of the value of the equation (1) and (2) in the equation (3). Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}{\frac{c}{a}}& = &{ - \frac{b}{a}}\end{array}\]
So
\[ \Rightarrow \begin{array}{*{20}{c}}b& = &{ - c}\end{array}\]
Therefore, the value will be \[b = - c\]
Hence, the correct option is B.

Note: In this question, it is important to note that when the roots are decreased by the \[1\] and then the new roots of the equation will be \[\alpha - 1\]and \[\beta - 1\] the product of these roots will give the condition that will be satisfied by the sum and the product of the roots of the previous equation.