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

Question

Nathan Axel · Accepted Answer

undefined To solve the quadratic equation $x^2 + 6x + 8 = 0$ using the quadratic formula:

1. Identify the coefficients $a$, $b$, and $c$:
   $a = 1$, $b = 6$, and $c = 8$.

2. Substitute these values into the quadratic formula:
   $$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

3. Plug in the values of $a$, $b$, and $c$:
   $$x = \frac{{-6 \pm \sqrt{{6^2 - 4(1)(8)}}}}{{2(1)}}$$

4. Simplify inside the square root:
   $$x = \frac{{-6 \pm \sqrt{{36 - 32}}}}{{2}}$$
   $$x = \frac{{-6 \pm \sqrt{{4}}}}{{2}}$$
   $$x = \frac{{-6 \pm 2}}{{2}}$$

5. Perform the calculations:
   $$x_1 = \frac{{-6 + 2}}{{2}} = \frac{{-4}}{{2}} = -2$$
   $$x_2 = \frac{{-6 - 2}}{{2}} = \frac{{-8}}{{2}} = -4$$

6. Therefore, the solutions are $x = -2$ and $x = -4$.

Eva Abramson · Answer

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

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#. Enter those figures into the quadratic formula to obtain: #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)#

Julia Adams · Answer

x = -2 or x = -4

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#