Question

Complete the steps for deriving the quadratic formula using the following equation.

Answer 1

I think this is it. hope this helps

Answer 2

Having obtained:

$x^2+ \frac{b}{a}x+( \frac{b}{2a} )^2+ \frac{c}{a}-( \frac{b}{2a} )^2=0$

the next thing to do is collect the first three terms and write them as the square of a binomial, and also collect the last two terms, writing each of them with a denominator equal to 4a^2

We have:

$(x+\frac{b}{2a} )^2+( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )=0$.

Then, we take $( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )$ to the right side and write these two terms as one:


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


Next, we take the square root of both sides, which has been shown in the solution.


Next, we have to take b/2a to the right hand side as -(b/2a), and removing the square in the denominator of the right hand side expression:

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



Answer: the steps to complete in the boxes are :

 
$(x+\frac{b}{2a} )^2+( \frac{4ac}{4a^2}-\frac{b^2}{4a^2} )=0$.

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

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