Question

Consider the first five steps of the derivation of the Quadratic Formula.

Answer 1

We want to derive the quadratic formula from the following expression

`x^2+(b)/(a)x+(b^2)/(4a^2)=-(c)/(a)+(b^2)/(4a^2)`

The next step would be rewrite the left side of the equation as the square of a sum, and combine the two terms on the right side of the equation on a single fraction.

Let's start by rewriting the left side.

When we expand the square of a binomial, we have

`(m+n)^2=m^2+2mn+n^2`

In our expression, we have

`x^2+(b)/(a)x+(b^2)/(4a^2)=x^2+(b)/(a)x+((b)/(2a))^2`

Comparing our expression with the expansion of a squared binomial, we have

`\begin{cases}m=x \\ n=(b)/(2a)\end{cases}\Rightarrow x^2+(b)/(a)x+((b)/(2a))^2=(x+(b)/(2a))^2`

Then, our original expression can be rewritten as

`\begin{gathered} x^2+(b)/(a)x+(b^2)/(4a^2)=-(c)/(a)+(b^2)/(4a^2) \\ (x+(b)/(2a))^2=-(c)/(a)+(b^2)/(4a^2) \end{gathered}`

And finally, by combining the terms on the right side, we have

`\begin{gathered} (x+(b)/(2a))^2=-(c)/(a)+(b^2)/(4a^2) \\ (x+(b)/(2a))^2=-(c)/(a)\cdot(4a)/(4a)+(b^2)/(4a^2) \\ (x+(b)/(2a))^2=-(4ac)/(4a^2)+(b^2)/(4a^2) \\ (x+(b)/(2a))^2=(-4ac+b^2)/(4a^2) \\ (x+(b)/(2a))^2=\frac{b^2-4ac^{}}{4a^2} \end{gathered}`

This is the next step. The answer is option C.