Question

Help with math plz! Giving brainlist

Answer 1

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

Answer 2

This is no 7. am solving the rest