How do you solve using completing the square method #x^2+3x+1=0#?

2 Answers

Answer:

The roots are

#x_1=(-3+sqrt5)/2# and #x_1=(-3-sqrt5)/2#

Explanation:

From the given #x^2+3x+1=0#, we can see that the coefficient of x^2 is already 1, so we can begin with the coefficient of x which is 3.

The 3 will have to be divided by 2 then the result should be squared and the final result is #color(red)(9/4)#. This number will be added and subtracted in the equation on one side.

#x^2+3x+1=0#

#x^2+3x+9/4-9/4+1=0#

The first 3 terms now will form a PST-perfect square trinomial.

#x^2+3x+9/4-9/4+1=0#

#(x^2+3x+9/4)-9/4+1=0#

this #(x^2+3x+9/4)# is equivalent to #(x+3/2)^2#

So, the equation becomes

#(x+3/2)^2-9/4+1=0#

simplify

#(x+3/2)^2-5/4=0#

transpose the 5/4 to the right side

#(x+3/2)^2=5/4#

Extract the square root of both sides of the equation

#sqrt((x+3/2)^2)=sqrt(5/4)#

#x+3/2=+-sqrt5/2#

#x=-3/2+-sqrt5/2#

The roots are

#x_1=(-3+sqrt5)/2# and #x_1=(-3-sqrt5)/2#

God bless....I hope the explanation is useful.

Mar 26, 2016

Answer:

#x=(sqrt5-3)/2# or #x=(-sqrt5-3)/2#

Explanation:

Given equation is #x^2+3x+1=0#

#x^2+3x=-1# -----------------------(1)

#Third term = (1/2 xx coefficient of x)^2#

#Third term = (1/2 xx 3)^2#

#Third term = (3/2)^2#

#Third term = 9/4#

Add #9/4# to both sides of equation (1)

#x^2+3x+9/4=-1+9/4#

#x^2+3x+9/4=(-4+9)/4#

#x^2+3x+9/4=5/4#

#(x+3/2)^2= 5/4#

#x+3/2=sqrt( 5/4)#

#x+3/2=+-sqrt( 5)/2#

#x=+-sqrt5/2-3/2#

#x=sqrt5/2-3/2# or #x=-sqrt5/2-3/2#

#x=(sqrt5-3)/2# or #x=(-sqrt5-3)/2#

#Solution set={(sqrt5-3)/2,(-sqrt5-3)/2}#