Question

Solve the following inequality: x^2-8x + 7>0

Answer 1

Answer:

The solution lie in  $(-\infty,1)\cup(7,\infty)$

Step-by-step explanation:

Given : Inequality $x^2-8x+7>0$

To find : Solve the inequality?

Solution :

First we convert the inequality into equation,

$x^2-8x+7=0$

Solving by middle term split,

$x^2-7x-x+7=0$

$x(x-7)-1(x-7)=0$

$(x-7)(x-1)=0$

$x=7,1$

Use each root to create test intervals,

$x$

For x<1, let x=0

$0^2-8(0)+7>0$

$0-0+7>0$

$7>0$

True.

For 1<x<7, let x=3

$3^2-8(3)+7>0$

$9-24+7>0$

$-8>0$

False.

For x>7, let x=8

$8^2-8(8)+7>0$

$64-64+7>0$

$7>0$

True.

Therefore, The inequality form is x<1 or x>7.

The interval notation is $(-\infty,1)\cup(7,\infty)$