Question

in the derivation of the quadratic formula by completing the square, the equation (left(x + \frac{b}{2a}
ight)^2 = \frac{-4ac + b^2}{4a^2}) is created by forming a perfect square trinomial. what is the result of applying the square root property of equality to this equation?

(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{-4ac + b^2}{4a^2})

(x + \frac{b}{2a} = \frac{pm sqrt{b^2 - 4ac}}{2a})

(x + \frac{b}{2a} = \frac{b pm sqrt{-4ac}}{2a})

(left(x + \frac{b}{2a}
ight)^2 = left(\frac{-4ac + b^2}{4a^2}
ight)^2)

Explanation:

Step1: Recall square root property

The square root property states that if $y^2 = z$, then $y = \pm\sqrt{z}$.

Step2: Apply to given equation

Let $y = x+\frac{b}{2a}$ and $z=\frac{-4ac+b^2}{4a^2}$. Substitute into the property:
$$x+\frac{b}{2a} = \pm\sqrt{\frac{b^2-4ac}{4a^2}}$$

Step3: Simplify the square root

$\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\sqrt{b^2-4ac}}{2|a|}$, and for quadratic formula we use $\frac{\sqrt{b^2-4ac}}{2a}$ with $\pm$ accounting for sign:
$$x+\frac{b}{2a} = \frac{\pm\sqrt{b^2-4ac}}{2a}$$

Answer:

B. $x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$