How do you solve #x = sqrt(-3x + 10) # and find any extraneous solutions?

Question

3376 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-09T09:37:09

#x=-5" and "x=+2" "# are solutions to #x^2+3x-10=0#

Extraneous #ul("values")# of #x>10/3 #

For #AAx notin {-5,2}" " =>" " x^2+3x-10 != 0#

#color(blue)("Determine general solution")#

Square both sides

#x^2=-3x+10" "larr" you now have a quadratic"#

#x^2+3x-10=0#

Notice that #2xx5=10" and "5-2=+3#

Thus the factorization is#" "(x+5)(x-2)=0#

So #x=-5" and "x=+2# are general solutions
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine extraneous values of "x)#

For the number to remain in the #ul("'Real'")# domain you may can not permit the square root to be that of a negative number.

So #-3x+10>=0#

Multiply by (-1) giving

#+3x-10<=0#
Notice that the inequality sign has turned the other way. This happens when you multiply by (-1)

Add 10 to both sides

#3x<=10#

#=>x<=10/3# as a required condition

#=>x>10/3 # takes the solution out of the Real domain and into the Complex number domain