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Can you please prove the quadratic formula?

\(\frac{-b+-\sqrt{b^2 - 4ac}}{2a}\)

 Aug 29, 2022
 #1
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Hi NJColonial6,

Yes, here is one way to prove the Quadratic formula:
Given a Quadratic Equation:              \(ax^2+bx+c=0\)                                                 (\(a \ne 0\))

Now, let's divide both sides by \(a\):

\(x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0 \)

Now, we want to complete the square (If you want to know how, go to the end of this answer (*).)
\(x^2+\dfrac{b}{a}x+\dfrac{c}{a}+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}=0 \\ (x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2})+\dfrac{c}{a}-\dfrac{b^2}{4a^2}=0 \\ \text{Notice the expression between the brackets, can be rewritten as:} \\ (x+\dfrac{b}{2a})^2+\dfrac{c}{a}-\dfrac{b^2}{4a^2}=0\)

 

Now, let's make a common denominator for the other two fractions:

\((x+\dfrac{b}{2a})^2+\dfrac{4ac}{4a^2}-\dfrac{b^2}{4a^2}=0 \\ \text{Notice we multiplied the numerator and denominator of c/a by 4a}\)

Now, we simplify this and move the fraction to the right side:

\((x+\dfrac{b}{2a})^2+\dfrac{4ac-b^2}{4a^2}=0 \\ (x+\dfrac{b}{2a})^2=-\dfrac{4ac-b^2}{4a^2} \\ \text{Distribute the negative on the right : } \\ (x+\dfrac{b}{2a})^2=\dfrac{b^2-4ac}{4a^2}\)

Now, take the square root of both sides:

\(\sqrt{(x+\dfrac{b}{2a})^2}=\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}=\dfrac{\pm\sqrt{b^2-4ac}}{\sqrt{4a^2}} \)

 

Simplify:

 

\(x+\dfrac{b}{2a}=\dfrac{\pm\sqrt{b^2-4ac}}{2a}\)

Solve for x:

\(x=-\dfrac{b}{2a}+\dfrac{\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Thus, we have proved the Quadratic formula.

 

 

*What is completing the square? 

Here:

If we have: \(x^2+bx+c=0\), we wish to write this as: \((x+k)^2+h\) for some constants k and h.

What we do is, we equate these two:

\(x^2+bx+c=(x+k)^2+h \\ \text{Then, expand the right-hand side:} \\ x^2+bx+c=x^2+2kx+k^2+h \\ \text{Equate coefficients:} \\ 1=1 \\ b=2k \implies k=\dfrac{b}{2} \\ c=k^2+h \implies h=c-k^2\)

So, we if wanted to complete the square of: 

\(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\)     into: \((x+k)^2+h\)

Our k is : \(k=\dfrac{b}{a} \div 2 =\dfrac{b}{2a}\)

The h is: \(h=\dfrac{c}{a}-(\dfrac{b}{2a})^2=\dfrac{c}{a}-\dfrac{b^2}{4a^2}\) as we found above.

 

 

I hope this helps!

 Aug 29, 2022
 #2
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Thank you so much geust! This helps a lot, thank you so much!

NJColonial6  Aug 29, 2022
 #3
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You are welcome :)

Guest Aug 29, 2022

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