How do you solve #x^2+8x+15=0 # using the quadratic formula?

2 Answers
Mar 20, 2018

Answer:

See a solution process below:

Explanation:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(1)# for #color(red)(a)#

#color(blue)(8)# for #color(blue)(b)#

#color(green)(15)# for #color(green)(c)# gives:

#x = (-color(blue)(8) +- sqrt(color(blue)(8)^2 - (4 * color(red)(1) * color(green)(15))))/(2 * color(red)(1))#

#x = (-color(blue)(8) +- sqrt(64 - 60))/2#

#x = (-color(blue)(8) - sqrt(4))/2# and #x = (-color(blue)(8) + sqrt(4))/2#

#x = (-color(blue)(8) - 2)/2# and #x = (-color(blue)(8) + 2)/2#

#x = (-10)/2# and #x = (-6)/2#

#x = -5# and #x = -3#

The Solution Set Is: #x = {-5, -3}#

Mar 20, 2018

Answer:

#x=-5# or #x=-3#

Explanation:

Quadratic formula gives the solution of a quadratic equation. For a quaddratic equation #ax^2+bx+c=0#, the roots given by quadratic formula are #(-b+-sqrt(b^2-4ac))/(2a)#.

Hence for #x^2+8x+15=0#, we have #a=1#, #b-8# and #c=15# and hence solution is

#(-8+-sqrt(8^2-4*1*15))/(2*1)#

= #(-8+-sqrt(64-60))/2#

= #(-8+-sqrt4)/2#

= #(-8+-2)/2#

i.e. either #x=-5# or #x=-3#