x^2+10x+20>4 solve for x values

Expert Answers
hala718 eNotes educator| Certified Educator

x^2+10x+20>4

First let us subtract 4 from both sides:

==> x^2+10x+16>0

Now factorize:

(x+2)(x+8)>0

==> x+2>0     and   x+8>0

==> x>-2    and   x>-8

Then x belongs to the interval (-2,infinity)

OR :

(x+8)<0     and     (x+2)<0

==> x<-8   and  x<-2

==> x belongs to the interval (-infinity, -8)

The x = (-inf,-8)U(-2,inf.)

OR x= R-[-8,-2]

neela | Student

To solve:x^2+10x+20 >4.

Solution:

The solution of an equation or an inequality is unaffected if we add or subtract equals on both sides . So we subtract 4  from both sides:

x^2+10x+20-4 > 4-4=0

x^2+10x+16>0. The LHS  is a quadratic expression which is greater than zero. Factorising the left we get:

(x+8)(x+2) > 0. So both factors  , x+8 and x+2 must be positive or negative in order that their product is positive or greater than zero.This is possible when  x > -2. Or x <-8. So, in the interval notation we write this : x  belongs to (-infinity,-8)  U (-2 , infinity)