Solve by completing the square: x^2+6x+4=0 (show work)?

x^2 + 6x + 4 = 0
x^2 + 6x = -4
x^2 + 3x + 3x = -4
x^2 + 3x + 3x + 9 = -4 + 9
(x^2 + 3x) + (3x + 9) = 5
x(x + 3) + 3(x + 3) = 5
(x + 3)(x + 3) = 5
(x + 3)^2 = 5
x + 3 = ±√5
x = -3 ±√5

x^2+6x+4=0
(x + 6/2) ^2 - (6/2)^2 + 4 = 0
(x + 3)^2 -9 + 4 = 0
(x + 3)^2 - 5 = 0

we've been taught a diverse technique, %. whichever is least puzzling for you. c^2 + 4c - (the quantity here is a million/2 the middle time era, i.e. 4, then sqaured) - 7 (then - the comparable time era in the different brackets) = 0 c^2 + 4c + 4 -7 - 4 = 0 (c+2)^2 -7 - 4 = 0 (The time era interior the brackets is a million/2 the middle time era, providing you with a appropriate sq.) (c+2)^2 -11 = 0 (c + 2 + sqrt11)(c + 2 - sqrt11)=0 (Factorise the equation to get this, a distinction of two squares) c=-2 +- sqrt11 Use the null element regulation to get your answer.

Question Number 1 :
For this equation x^2 + 6*x + 4 = 0 , answer the following questions :
A. Find the roots using Quadratic Formula !
B. Use completing the square to find the root of the equation !

Answer Number 1 :
The equation x^2 + 6*x + 4 = 0 is already in a*x^2+b*x+c=0 form.
In that form, we can easily derive that the value of a = 1, b = 6, c = 4.

1A. Find the roots using Quadratic Formula !
Use the formula,
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
As a = 1, b = 6 and c = 4,
we need to subtitute a,b,c in the abc formula, with thos values.
Which produce x1 = (-(6) + sqrt( (6)^2 - 4 * (1)*(4)))/(2*1) and x2 = (-(6) - sqrt( (6)^2 - 4 * (1)*(4)))/(2*1)
Which make x1 = ( -6 + sqrt( 36-16))/(2) and x2 = ( -6 - sqrt( 36-16))/(2)
Which is the same as x1 = ( -6 + sqrt( 20))/(2) and x2 = ( -6 - sqrt( 20))/(2)
So we get x1 = ( -6 + 4.47213595499958 )/(2) and x2 = ( -6 - 4.47213595499958 )/(2)
So we have the answers x1 = -0.76393202250021 and x2 = -5.23606797749979

1B. Use completing the square to find the root of the equation !
x^2 + 6*x + 4 = 0 ,divide both side with 1
Which result in x^2 + 6*x + 4 = 0 ,
The coefficient of x is 6
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = 6/2 = 3
Which means we can turn the equation into x^2 + 6*x + 9 - 5 = 0
Which is the same with ( x + 3 )^2 - 5 = 0
So we will get (( x + 3 ) - 2.23606797749979 ) * (( x + 3 ) + 2.23606797749979 ) = 0
By using the associative law we get ( x + 3 - 2.23606797749979 ) * ( x + 3 + 2.23606797749979 ) = 0
Just add up the constants in each brackets, and we get ( x + 0.76393202250021 ) * ( x + 5.23606797749979 ) = 0
The answers are x1 = -0.76393202250021 and x2 = -5.23606797749979

x² + 6x + 9 = 9 - 4
(x + 3)² = 5
x + 3 = ±√5
x = - 3 ± √5

x² + 6x + 4 = 0
The trinomial is not factorable. We have to use the quadratic formula.
x = {-6 ± √[6² - 4(1)(4)]}/2
x = {-6 ± √20}/2
x = {-6 ± 2√5}/2
x = -3 ± √5 = -5.236067978, -0.763932022

That's all the work I can show.
Happy Holidays!

To solve by completing the square convert the equation into a form that can be factored as a square.

For example if you expand (x+b)^2 you get

x^2 + 2bx +b^2

That is the form you want your equation in.

In your case you start with

x^2+6x+4=0

Now add 5 to both sides:

x^2+6x+9 = 5

The general rule is that you want the last term before the equals sign to be the square of half the coefficient of the receding term. Inthis case the coefficent of x is 6 and half of 6 squared is 9. To get the last term to be 9 you have add 5 to each side.

Now it is in the required form and you can now say

(x+3)^2 = 5

and

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

x^2+6x+4=0
x^2+2x*3+3^2-3^2+4=0
(x+3)^2=5
x+3=+-sqrt(5)
x=-3+-sqrt(5)
I hope that you understand this, and good luck!

Here's how:

First transpose the constant to the other side of the equation

X^2+6X+4=0
X^2 +6X= - 4

Complete the square

X^2 + +6X + { ? } = -4 + { ? } (equals added to equals, sum are equals

The probable square is 9, 3X3=9 place inside { }

X^2 + 6X + 9 = -4 + 9

Factor the quadratic equation and extract the square root of the constant


(X+3)(X+3) = 5
(X+3) = (5)^(1/2)
X= the positive value of square root of 5 minus 3
X=the negative value of square root of 5 minus 3

x^2+6x+4=0 is re-written as

x^2+ 2(3x) + 9 - 9 + 4 = 0 ( If you simply this, u get what's on top )

which becomes

(x+3)^2 - 5 = 0. ( because a^2 + 2ab + b^2 = (a+b)^2 )

Done.