Question

Extraneous solutions
http://awesomescreenshot.com/0782bazb9c

Answer 1

\[\sqrt{3x+28}-8=x\]
\[\sqrt{3x+28}=x+8\]
squaring both sides
\[3x+28=\left( x+8 \right)^2\]
\[3x+28=x^2+16x+64~or~x^2+16x-3x+64-28=0\]
\[x^2+13x+36=0\]
\[x^2+9x+4x+36=0\]
x(x+9)+4(x+9)=0
(x+9)(x+4)=0
now you can solve.

Answer 2

how do I solve o.o

Answer 3

either x+4=0,x=-4
or x+9=0,x=-9

Answer 4

Am I supposed to plug those in as x

Answer 5

when x=-4
\[\sqrt{3*-4+28}-8=-4,4-8=-4,-4=-4\]
whenx=-9
\[\sqrt{3*-9+28}-8=-9,1-8=-9,-7=-9,which is \not true.\]
Hence x=-9 is an extraneous root.

Answer 6

would positive 4 be an extraneous root

Answer 7

no, only x=-9

Answer 8

Ohhh understood.. appreciate it