Question

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

Answer 1

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.

Answer 2

`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}`