Question

Solve for the zeros of the quadratic function f(x) = 9x² + 6x + 1.

Answer 1

Answer:

$\large\boxed{x=-\dfrac{1}{3}}$

Step-by-step explanation:

$f(x)=9x^2+6x+1\\\\\text{The zeros are for}\ f(x)=0\\\\9x^2+6x+1=0\\\\9x^2+3x+3x+1=0\\\\3x(3x+1)+1(3x+1)=0\\\\(3x+1)(3x+1)=0\\\\(3x+1)^2=0\iff3x+1=0\qquad\text{subtract 1 from both sides}\\\\3x=-1\qquad\text{divide both sides by 3}\\\\x=-\dfrac{1}{3}$