Answer :
Answer:
The solution lie in [tex](-\infty,1)\cup(7,\infty)[/tex]
Step-by-step explanation:
Given : Inequality [tex]x^2-8x+7>0[/tex]
To find : Solve the inequality?
Solution :
First we convert the inequality into equation,
[tex]x^2-8x+7=0[/tex]
Solving by middle term split,
[tex]x^2-7x-x+7=0[/tex]
[tex]x(x-7)-1(x-7)=0[/tex]
[tex](x-7)(x-1)=0[/tex]
[tex]x=7,1[/tex]
Use each root to create test intervals,
[tex]x<1\\1<x<7\\x>7[/tex]
For x<1, let x=0
[tex]0^2-8(0)+7>0[/tex]
[tex]0-0+7>0[/tex]
[tex]7>0[/tex]
True.
For 1<x<7, let x=3
[tex]3^2-8(3)+7>0[/tex]
[tex]9-24+7>0[/tex]
[tex]-8>0[/tex]
False.
For x>7, let x=8
[tex]8^2-8(8)+7>0[/tex]
[tex]64-64+7>0[/tex]
[tex]7>0[/tex]
True.
Therefore, The inequality form is x<1 or x>7.
The interval notation is [tex](-\infty,1)\cup(7,\infty)[/tex]