SOLUTION
THE ERRORS
1. STEP 2 : 8 shouldn't have been added to bothsides
2. STEP 3: The expansion of the LHS expression wouldn't give LHS of step 2
THE SOLUTION BREAKDOWN
[tex]\begin{gathered} x^2+8x=0 \\ multiply\text{ bithsides by \lparen}\frac{coefficient\text{ of x}}{2})^2 \\ (\frac{1}{2}\times8)^2=\text{ \lparen4\rparen}^2 \\ (4)^2\text{ will be added to bothsides } \\ x^2+8x+(4)^2=0+4^2 \\ (x+4)^2=16 \\ observe\text{ that the expansion of \lparen x+4\rparen}^2=x^2+8x+(4)^2 \\ take\text{ square root of bothsides } \\ \sqrt{(x+4)^2}=\sqrt{16} \\ the\text{ square cancels out the square root on the LHS } \\ x+4=\pm4 \\ x=-4\pm4 \\ x_1=-4+4=0 \\ x_2=-4-4=-8 \end{gathered}[/tex]
The final answers are x=0 , x=-8
