Some of the steps in the derivation of the quadratic formula are shown.
Which best explains why the expression $ \pm \sqrt{b^{2}-4 a c} $ $ \operatorname{step} 4: \frac{-4 a c+b^{2}}{4 a}=a\left(x+\frac{b}{2 a}\right)^{2} $ cannot be rewritten as $ b \pm \sqrt{-4 \alpha} $ during the next step?
$ \operatorname{step} 5:\left(\frac{1}{a}\right) \frac{b^{2}-4 a c}{4 a}=\left(\frac{1}{a}\right) a\left(x+\frac{b}{2 a}\right)^{2} $
Negative values, like $ -4 a c $, do not have a square root.

The $ \pm $ symbol prevents the square root from being evaluated.
Step $ 6: \frac{b^{2}-4 a c}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2} $
The square root of terms separated by addition and subtraction cannot be calculated individually.
Step $ 7: \frac{\pm \sqrt{b^{2}-4 a c}}{2 a}=x+\frac{b}{2 a} $ The entire term $ b^{2}-4 a c $ must be divided by $ 2 a $ before its square root can be determined.

Question

Some of the steps in the derivation of the quadratic formula are shown.
Which best explains why the expression $ \pm \sqrt{b^{2}-4 a c} $ $ \operatorname{step} 4: \frac{-4 a c+b^{2}}{4 a}=a\left(x+\frac{b}{2 a}\right)^{2} $ cannot be rewritten as $ b \pm \sqrt{-4 \alpha} $ during the next step?
$ \operatorname{step} 5:\left(\frac{1}{a}\right) \frac{b^{2}-4 a c}{4 a}=\left(\frac{1}{a}\right) a\left(x+\frac{b}{2 a}\right)^{2} $
Negative values, like $ -4 a c $, do not have a square root.

The $ \pm $ symbol prevents the square root from being evaluated.
Step $ 6: \frac{b^{2}-4 a c}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2} $
The square root of terms separated by addition and subtraction cannot be calculated individually.
Step $ 7: \frac{\pm \sqrt{b^{2}-4 a c}}{2 a}=x+\frac{b}{2 a} $ The entire term $ b^{2}-4 a c $ must be divided by $ 2 a $ before its square root can be determined.

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