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Show work and/or describe how the expression for the completing the square method and the expression associated with the quadratic formula are equivalent.

Sagot :

Given a general quadratic expression:

[tex]ax^2+bx+c=0[/tex]

firs, lets divide both sides of the equation by 'a' :

[tex]\begin{gathered} (\frac{1}{a})(ax^2+bx+c=0)^{} \\ \Rightarrow\frac{a}{a}x^2+\frac{b}{a}x+\frac{c}{a}=0 \\ \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0 \end{gathered}[/tex]

next, we can move the term c/a to the right side of the equation:

[tex]\begin{gathered} x^2+\frac{b}{a}x+\frac{c}{a}=0 \\ \Rightarrow x^2+\frac{b}{a}x=-\frac{c}{a} \end{gathered}[/tex]

now we are ready to complete the square on the left side. What we have to do, is to take the constant that is multiplying x (in this case,b/a), and first, we divide it by 2, and then elevate to the square the result:

[tex]\begin{gathered} \frac{b}{a}\frac{\cdot}{\cdot}2=\frac{b}{2a} \\ \Rightarrow(\frac{b}{2a})^2=\frac{b^2}{4a^2} \end{gathered}[/tex]

then, adding this number on both sides of the equation, we get:

[tex]x^2+\frac{b}{a}x+\frac{b^2}{4a}=-\frac{c}{a}+\frac{b^2}{4a^2}[/tex]

which we can write like this:

[tex](x+\frac{b}{2a})^2=\frac{-4ac+b^2}{4a^2}_{}[/tex]

applying the square root on both sides,we get the following:

[tex]\begin{gathered} \sqrt[]{(x+\frac{b}{2a})^2}=\sqrt[]{\frac{b^2-4ac}{4a^2}}=\pm\frac{\sqrt[]{b^2_{}-4ac}}{2a} \\ \Rightarrow x+\frac{b}{2a}=\pm\frac{\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]

finally, we can solve for x:

[tex]\begin{gathered} x+\frac{b}{2a}=\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ \Rightarrow x=-\frac{b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]

as we can see, if we have a general quadratic equation, we can us the completing the square method to deduce the quadratic formula

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