extranious solution is mostly when you take the square root of negative number
what you want to do is math stuff
translate
remember that \(x^ \frac{1}{2} = \sqrt{x}\)
so
\(x^ \frac{3}{2}=x \sqrt{x}\)
7.
\((x+1) \sqrt{x+1} -2=25\)
add 2 to both sides
\((x+1) \sqrt{x+1}=27\)
square both sides
\((x+1)^3=729\)
cube root both sides
x+1=9
minus 1
x=8
8.
square both sides
\(3x+7=x^2-2x+1\)
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
\(2x+6=4 \sqrt{x+1}+x+5\)
minus (x+5) from both sides
\(x+1=4 \sqrt{x+1}\)
square both sides
\(x^2+2x+1=16x+16\)
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
