Why does the equation for roots in a quadratic equation actually work? Please explain. I am curious to know.

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 ...

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.

step-by-step explanation:

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