When solving an equation in the form #ax^2=c# by taking square root how many solutions will there be?

1 Answer
Jan 12, 2017

Answer:

There can be #0#, #1#, #2# or infinitely many.

Explanation:

Case #bb(a=c=0)#

If #a=c=0# then any value of #x# will satisfy the equation, so there will be an infinite number of solutions.

#color(white)()#
Case #bb(a=0, c!= 0)#

If #a=0# and #c!=0# then the left hand side of the equation will always be #0# and the right hand side non-zero. So there is no value of #x# which will satisfy the equation.

#color(white)()#
Case #bb(a!=0, c=0)#

If #a != 0# and #c=0# then there is one solution, namely #x=0#.

#color(white)()#
Case #bb(a > 0, c > 0)# or #bb(a < 0, c < 0)#

If #a# and #c# are both non-zero and have the same sign, then there are two Real values of #x# which satisfy the equation, namely #x = +-sqrt(c/a)#

#color(white)()#
Case #bb(a > 0, c < 0)# or #bb(a < 0, c > 0)#

If #a# and #c# are both non-zero but of opposite sign, then there are no Real values of #x# which satisfy the equation. If you allow Complex solutions, then there are two solutions, namely #x = +-i sqrt(-c/a)#