Answer :
We are given the following quadratic inequality.
[tex]x^2+7\ge8x[/tex]Let us convert it into quadratic form and then factorize the equation
[tex]\begin{gathered} x^2+7\ge8x \\ x^2-8x+7\ge0 \end{gathered}[/tex]So now we need two numbers such that their sum is equal to -8 and their product is equal to 7.
How about -7 and -1 ?
Sum = -7 - 1 = -8
Product = -7*-1 = 7
[tex]\begin{gathered} x^2-8x+7\ge0 \\ (x-7)(x-1)\ge0 \end{gathered}[/tex]Now there are few possible cases
[tex]\begin{gathered} (x-7)\ge0\quad and\quad (x-1)\ge0 \\ x\ge7\quad and\quad x\ge1 \end{gathered}[/tex]Since x => 1 meets the requirement of x => 7 so we take x => 1 form here.
The other possible case is
[tex]\begin{gathered} (x-7)\le0\quad and\quad (x-1)\le0 \\ x\le7\quad and\quad x\le1 \end{gathered}[/tex]Since x <= 7 meets the requirement of x <= 1 so we take x <= 7 form here.
So the solution becomes
[tex]\begin{gathered} x\ge1\quad and\quad x\le7 \\ 1\le x\le7 \end{gathered}[/tex]Finally, the solution in interval notation is written as
[tex]\lbrack1,7\rbrack[/tex]We use brackets for closed intervals (less, greater or equal than cases)
We use parenthesis for open intervals (less, greater than cases)