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
