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Question

Solve this equation

x^2 + 6x + 8 = 0

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Answer 1

There are two ways to solve this.

One is to figure out the factoring. Since the coefficient of x (+6) and the constant (+8) are positive, it would be factored as (x+a)(x+b) where the sum a+b is 6, and product a*b is 8. That works for the obvious choices for factors of eight (4*2 = 8, 4+2 = 6). The equation is then factored as :

(x + 2)(x + 4) = 0

That product equals zero when either term is zero, which means that the two solutions are:

x+2 = 0 --> x = -2
x+4 = 0 --> x= -4

The other way is to use the quadratic equation, which solves ax^2 + bx + c = 0 for any values of a, b, and c (1, 6, and 8 in this case):

x = ( -b +/- sqrt(b^2 - 4ac) ) /2a
x = ( -6 +/- sqrt(6^2 - 4*1*8) ) / (2*1)
x = ( -6 +/- sqrt(36 - 32) ) /2
x = ( -6 +/- sqrt(4) ) / 2
x = ( -6 +/- 2 ) / 2
x = -3 +/- 1

x = (-3 + 1) = -2
x = (-3 - 1) = -4

Factoring is easier for simple problems like this one, but you should use whatever solution your teacher is expecting you to learn for this particular homework.

Answer 2

x^2 + 6x + 8 = 0
=> x^2 + 4x + 2x +8 = 0
=> x(x + 4) + 2(x + 4) = 0
=> (x + 4)(x + 2) = 0
=> x + 4 =0 or x + 2 = 0
=> x = - 4 or x = -2

Answer 3

You can use the quadratic equation to solve it or you can just factor it to solve.

The solution will be of the form (x+a)(x+b)=0 so you have 2 unknowns and 2 equations

ab=8
a+b=6

4*2=8 and 4+2 =6 so a=4 and b=2

(x+4)(x+2)=0

This is 0 only when either x+4=0 or when x+2=0 so either

x=-4 or x=-2 will satisfy the equaiton.

The quadratic equation is of the form ax^2+bx+c=0

Here a=1, b=6 and c=8

The general solution to this is x = [-b+/-sqrt(b^2-4ac)] / 2a

x = [-6 +/- sqrt(6^2 - 4*1*8)] / 2
x = [-6 +/- sqrt(36-32)]/2
x = [-6 +/- sqrt(4)]/2
x = [-6 +/- 2]/2

x=(-6+2)/2=-2 or x=(-6-2)/2=-4 as determined by factoring.

Answer 4

First: multiply the 1st & 3rd term to get 8. find two numbers that give you 8 when multiplied & 6 (2nd term) when added/subtracted. the numbers are (2 & 4). rewrite the expression with the new middle terms.

x^2 + 2x + 4x + 8 = 0

Sec: with 4 terms - group "like" terms & factor both sets of parenthesis.

(x^2 + 2x) + (4x + 8) = 0
x(x + 2) + 4(x + 2) = 0
(x + 2)(x + 4) = 0

Third: set both parenthesis to equal "0" & solve for "x"

a. x + 2 = 0 > subtract 2 from both sides (when you move a term to the opposite side, always use the opposite sign).

x + 2 - 2 = 0 - 2
x = 0 - 2
x = -2

b. x+ 4 = 0 > subtract 4 from both sides.

x + 4 - 4 = 0 - 4
x = 0 - 4
x = - 4

Solution: -2, - 4

Answer 5

here
x*2+6x+8=0
(x+2)(x+4)=0
x+2=0 x+4=0
x=0-2 x=0-4
x=-2 x=-4

Answer 6

It's very hard to tell you how I get this, but its just experience from factoring
(x+2)(x+4)=0
x= -2 or -4
There is also a quadratic formula.
(-b+/-sqrt of b^2-4ac)/2a
b^2-4ac is called the discriminant.
It's useful for determining a graph.
But well you'll understand it's properties through studying graphs.