How do you solve #9x^2+6x+1=0# using completing the square?

Question

Julia Adams · Accepted Answer

Recognizing that #9x^2+6x+1# we get #x= -1/3# #9x^2+6x+1 = 0# is equivalent to: #color(white)("XXXX")##(3x+1)^2 = 0# #rArr# #color(white)("XXXX")##3x+1 = 0# #rArr# #color(white)("XXXX")##x = -1/3#

Isaiah Anthony · Answer

undefined To solve the quadratic equation $9x^2 + 6x + 1 = 0$ using completing the square:

1. Move the constant term to the other side: $9x^2 + 6x = -1$.
2. Divide all terms by the coefficient of $x^2$ to make the leading coefficient 1: $x^2 + \frac{2}{3}x = -\frac{1}{9}$.
3. To complete the square, take half of the coefficient of $x$, square it, and add it to both sides of the equation: 
   $x^2 + \frac{2}{3}x + \left(\frac{1}{3}\right)^2 = -\frac{1}{9} + \left(\frac{1}{3}\right)^2$.
4. Simplify: $x^2 + \frac{2}{3}x + \frac{1}{9} = -\frac{1}{9} + \frac{1}{9}$.
5. Factor the left side and simplify the right side: $(x + \frac{1}{3})^2 = 0$.
6. Take the square root of both sides: $x + \frac{1}{3} = 0$.
7. Solve for $x$: $x = -\frac{1}{3}$.

So, the solution to the quadratic equation $9x^2 + 6x + 1 = 0$ using completing the square is $x = -\frac{1}{3}$.