Answer:
[tex]\frac{b^2 - 4ac}{4a^2}[/tex]
Step-by-step explanation:
Given
See attachment for complete question
Required:
Complete step 6
At step 5, we have:
[tex](x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}[/tex]
Take LCM
[tex](x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}[/tex]
Take square roots of both sides to get step 6
[tex]x + \frac{b}{2a} = \±\sqrt{\frac{b^2 - 4ac}{4a^2}}[/tex]
Hence, the missing radicand is: [tex]\frac{b^2 - 4ac}{4a^2}[/tex]
