How do you solve #-x ^ { 2} + 4x + 3\geq 0#?

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Answer 1 · 2017-09-29T11:25:18

Solution: #(2 -sqrt7) < x < (2+sqrt7) # or in interval notation,
#[2 -sqrt7 , 2 +sqrt7]#

#-x^2+4x+3 >= 0 or x^2-4x-3 <=0#

We will find zeros of #x^2-4x-3#

#x^2-4x-3 =0 or x^2-4x = 3 or x^2-4x+4 = 3+4 # or

#(x-2)^2 = 7 or x-2 = +- sqrt 7 or x= 2 +- sqrt 7#

#:. x=2+sqrt7 ~~ 4.65(2dp) ,x= 2 -sqrt7 ~~ -0.65 (2dp)#

# x^2-4x-3 <=0 or (x+0.65)( x-4.65) <=0 #

Critical points are # x= -0.65 , x=4.65#

#x^2-4x-3=0#,when #x= 2 -sqrt7 and x=2+sqrt7#

Sign chart:

When #x < -0.65# sign of #(x+0.65)( x-4.65)#

is #( -*- =+) ; > 0#

When #-0.65 < x < 4.65# sign of #(x+0.65)( x-4.65)#

is #( +*- =-) ; < 0#

When #x > 4.65# sign of #(x+0.65)( x-4.65)#

is #( +*+=+) ; >0 #

Solution: #(2 -sqrt7) < x < (2+sqrt7) # or in interval notation,

#[2 -sqrt7 , 2 +sqrt7]#

Solution: ( In decimal): #-0.65 < x < 4.65 # or in interval notation,

#[-0.65 , 4.65]# [Ans]