Question

Extraneous solutions

Answer 1

http://prntscr.com/l0n1v6

Answer 2

x - a = sqrt(bx+c)

it wants x = 7 as a solution, so

7 - a = sqrt(7b + c)

through trial and error I found that

a = 1
b = 3
and
c = 15 will give an extraneous solution

plugging them in gives us

7 - 1 = sqrt(7*3 + 15)
6 = 6

but going back and trying to solve the equation manually

x - a = sqrt(bx+c)

squaring both sides

(x-a)^2 = bx + c
(x-1)^2 = 3x + 15
x^2 - 2x + 1 = 3x + 15
solving this algebraically gives us x = 7 and x = -2, but the x = -2 is extraneous (try plugging this back into the original equation and you will see)

Answer 3

for b) bx + c must be greater than or equal to 0, otherwise you have a negative number under the radical which will result in an imaginary #

Answer 4

why did you have to go back and "solve the equation manually"

Answer 5

to find the extraneous solution