Explanation:
The formula for the solution of a quadratic is based on the the solution method "completing the square." The idea there is that the solution to ...
(x -p)² = q
can be found by taking square roots:
x -p = ±√q
so the values of x will be ...
x = p ±√q . . . . . . add p to both sides
Considering the above, take a look at (x -p)². When we write it out, we have ...
(x -p)² = x² -2px +p²
So, in the equation ...
(x -p)² = q
we can replace the left side to get ...
x² -2px +p² = q
x² -2px +(p² -q) = 0 . . . . . . subtract q to get standard form
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Compare this to the general quadratic:
ax² +bx +c = 0
Dividing that by 'a' gives the equivalent monic quadratic ...
x² +(b/a)x +(c/a) = 0
When we compare this to the equation with p and q, we see that ...
-2p = b/a . . . . . . . . coefficient of x
p² -q = c/a . . . . . . . constant
Solving the first of these for p gives ...
p = -b/(2a)
and using this in the second of these, we can find q to be ...
q = p² -c/a = (-b/(2a))² -c/a
q = (b² -4ac)/(4a²) . . . . combining the terms into a single fraction
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Now that we have p and q in terms of a, b, c, we can show the solution to the quadratic to be ...
x = p ± √q
x = -b/(2a) ± √((b² -4ac)/(4a²))
This can be expressed over a single denominator as ...
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
This is the familiar form of the quadratic formula.
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Additional comment
Here, we have arrived at the formula by sort of working backwards from a kind of vertex form of the quadratic. ((x-p)²=q has vertex (p, -q)). The more usual derivation uses the "complete the square" algorithm steps starting with ax²+bx+c=0. The intention is to show that the values of x given by the formula are actually the values of x that solve the equation, which is what you wanted to see.
