Question

which equation shows x^2+6x-4=0 rewritten by completing the squarea) (x+3)^2=36b) (x+3)^2=4c) (x+3)^2=9d) (x+3)^2=13

Answer 1

Solution

Step 1

Write the equation:

$x^2\text{ + 6x - 4 = 0}$

Step 2:

Rewrite the equation:

$x^2\text{ + 6x = 4}$

Step 3

$\begin{gathered} Add\text{ }\frac{b^2}{4a\text{ }}\text{ to both sides to get a perfect square.} \\ \text{a = 1, b = 6} \\ \frac{b^2}{4a}\text{ = }\frac{6^2}{4\times1}\text{ = }\frac{36}{4}\text{ = 9} \end{gathered}$$\begin{gathered} x^2\text{ + 6x + 9 = 4 + 9} \\ Add\text{ similar terms:} \\ (x\text{ + 3\rparen}^2\text{ = 13} \end{gathered}$

Final answer

$d)\text{ \lparen x + 3\rparen}^2\text{ = 13}$