How do you solve #\sqrt { 19- x } = x + 1#?

Question

899 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-18T13:01:20

Square both sides.

#(sqrt(19 - x))^2 = (x + 1)^2#

#19 - x = x^2 + 2x + 1#

#0 = x^2 + 3x - 18#

#0 = (x + 6)(x - 3)#

#x = -6 and 3#

However, #x= -6# is extraneous since it does not satisfy the original equation .

Hopefully this helps!

Answer 2 · 2016-10-18T13:06:49

#x=3#

#sqrt(19-x)=x+1#

Squaring both sides:
#rarr 19-x=x^2+2x+1#

Writing in standard form:
#rarr x^2+3x-18=0#

Factoring
#rarr (x+6)(x-3)=0#

#rArr#
#color(white)("XX"){: (x=-6,color(white)("XX")orcolor(white)("XX"),x=3) :}#

Since #sqrt("anything")# is taken to be the primary (i.e. non-negative) root, we can eliminate #x=-6# as being extraneous.