extranious solution is mostly when you take the square root of negative number
what you want to do is math stuff
translate
remember that [tex]x^ \frac{1}{2} = \sqrt{x} [/tex]
so
[tex]x^ \frac{3}{2}=x \sqrt{x} [/tex]
7.
[tex](x+1) \sqrt{x+1} -2=25[/tex]
add 2 to both sides
[tex](x+1) \sqrt{x+1}=27[/tex]
square both sides
[tex](x+1)^3=729[/tex]
cube root both sides
x+1=9
minus 1
x=8
8.
square both sides
[tex]3x+7=x^2-2x+1[/tex]
using math
0=x²-5x-6
x=6 or -1
if we do x=-1, we get
√4=-2, which is kind of true, but we normally find the principal root (posiitive0
-1 is extraious
x=6
9.
add √(x-1) to both sides then square both sides
[tex]2x+6=4 \sqrt{x+1}+x+5[/tex]
minus (x+5) from both sides
[tex]x+1=4 \sqrt{x+1} [/tex]
square both sides
[tex]x^2+2x+1=16x+16[/tex]
using math
x=-1 or 15
if we had x=-1, we would be having √(-1-1) which is sqrt of negative which is not allowed
x=15
