How do you solve #sqrt(9x+10)=x# and find any extraneous solutions?

Question

Genesis Allen · Accepted Answer

#x=10# or #x=-1#
(no extraneous solutions)

Given #color(white)("XXX")sqrt(9x+10)=x# Square both sides (this where an extraneous solution might be introduced) #color(white)("XXX")9x+10=x^2# Rearrange as a quadratic in standard form #color(white)("XXX")x^2-9x-10=0# Factor #color(white)("XXX")(x-10)(x+1)=0# Either #color(white)("XXX")(x-10)=0color(white)("XX")rarrcolor(white)("XX")x=10# or #color(white)("XXX")(x+1)=0color(white)("XX")rarrcolor(white)("XX")x=-1# Checking against original equation: #color(white)("XXX")#If #x=10# #color(white)("XXXXXXX")sqrt(9x+10)=sqrt(9(10)+10)=sqrt(100)=10=x# #color(white)("XXX")#This solution is valid. #color(white)("XXX")#If # x=-1# #color(white)("XXXXXX")sqrt(9x+10)=sqrt(9(-1)+10)=sqrt(1)=1=x# #color(white)("XXX")#This solution is valid

Eva Abramson · Answer

undefined To solve the equation sqrt(9x+10)=x and find any extraneous solutions, we can follow these steps:

1. Square both sides of the equation to eliminate the square root: (sqrt(9x+10))^2 = x^2
   This simplifies to 9x + 10 = x^2.

2. Rearrange the equation to form a quadratic equation: x^2 - 9x - 10 = 0.

3. Factorize the quadratic equation: (x - 10)(x + 1) = 0.

4. Set each factor equal to zero and solve for x:
   x - 10 = 0  -->  x = 10
   x + 1 = 0   -->  x = -1

5. Check for extraneous solutions by substituting the found values back into the original equation:
   For x = 10: sqrt(9(10) + 10) = 10
   Simplifying, we get sqrt(100) = 10, which is true.

For x = -1: sqrt(9(-1) + 10) = -1
   Simplifying, we get sqrt(1) = -1, which is not true.

Therefore, the solution to the equation sqrt(9x+10)=x is x = 10, and the extraneous solution is x = -1.