How do you solve #7x^2-6=57#?

4 Answers
Oct 3, 2017

Answer:

#x=+3 orx =-3#

Explanation:

There is no term in #x#, so you can solve for #x^2#

#7x^2 = 57+6" "larr# add 6 to both sides.

#7x^2 = 63" "larr div 7#

#x^2 =9#

#x= +-sqrt9#

#x =+-3#

Oct 3, 2017

Answer:

Solution: # x=3 , x = -3#

Explanation:

#7x^2 -6 = 57 or 7x^2 =57+6 or 7x^2 =63# or

#x^2 =63/7 or x^2 =9 or x = +- sqrt 9 or x = +-3#

Solution: # x=3 , x = -3# [Ans]

Oct 3, 2017

Answer:

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

Explanation:

First, we add #6# to both sides.
#7x^2-6=57#
#7x^2-6color(blue)+color(blue)6=57color(blue)+color(blue)6#
Simplify:
#7x^2=63#

Divide both sides by #7#

#(7x^2)/7=63/7#

Therefore, #x^2=9#

If #x^2=a#, then #x=sqrta# or #-sqrta#.

#9=3^2#
#x=sqrt(3^2)#
#x=3#

or

#x=-sqrt9#
#x=-sqrt(3^2)#
#x=-3#

Both ways work.

Oct 3, 2017

Answer:

#x=+-3#

Explanation:

If
#color(white)("XXX")7x^2-6=57#
then
#color(white)("XXX")7x^2=63# ...after adding #6# to both sides
and
#color(white)("XXX")x^2=9# ...after dividing both sides by #7#

Therefore
#color(white)("XXX")x=+3" or " x=-3# ...after taking the square root of both sides (since #(+-3)^2=9#