How do you solve #x^2 +6x +8 =0# using the quadratic formula?

2 Answers
May 8, 2018

Answer:

The answers are #x=-2# and #x=-4#.

Explanation:

To start, the quadratic formula is #x=(-bpmsqrt(b^2-4ac))/(2a)#

In this problem, #a = 1# (as the #x^2# term has no coefficient), #b=6#, and #c=8#.

Plug those values into the quadratic equation to get:

#x=(-6pmsqrt(6^2-4(1)(8)))/(2(1))#

Multiply #2*1# on the bottom of the fraction:

#x=(-6pmsqrt(6^2-4(1)(8)))/(2)#

Square #6# and multiply #4*1*8# within the square root:

#x=(-6pmsqrt(36-32))/(2)#

Subtract #36-32# inside the root:

#x=(-6pmsqrt(4))/(2)#

Solve for #sqrt(4)#

#x=(-6pm2)/(2)#

If the #pm# is positive, you get

#x=(-6+2)/(2)#, which simplifies to #x=(-4)/(2)#, or #color(red)(-2)#

If the #pm# is negative, you get

#x=(-6-2)/(2)#, which simplifies to #x=(-8)/(2)#, or #color(red)(-4)#

May 8, 2018

Answer:

x = -2 or x = -4

Explanation:

The quadratic formula looks like
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

You have...
#ax^2 + bx + c = 0#
and your equation...
#x^2 + 6x +8 = 0#

With that you can do...
#x^2 +6x + 8 = 0#
#a = 1#
#b = 6#
#c = 8#

Then you substitute what you have into the quadratic formula
When you do that you get...
#x = (-(6) +- sqrt((6)^2 - 4(1)(8)))/(2(a))#

The '#+-#' means that your going to have 2 answers like x = this or that so, you can just solve the whole equation using '#+#' first then use '#-#' next.

when you do that you'll get an answer of #x = -2 or -4#