Question

I need help with this question please. This is non- graded.

Answer 1

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

`\begin{gathered} x^2+8x=0 \\ multiply\text{ bithsides by \lparen}\frac{coefficient\text{ of x}}{2})^2 \\ ((1)/(2)*8)^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 } \\ √((x+4)^2)=√(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}`

The final answers are x=0 , x=-8