Math

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how do you prove the quadratic equation?

  • Math -

    the proof is simply to complete the square for a generalised quadratic equation. Like this:
    ax2(xsquared) + bx + c = 0
    Take 'a' outside:
    a[x2 + bx/a + c/a] = 0
    Divide through by 'a':
    x2 + bx/a + c/a = 0
    Complete the square:
    (x + b/2a)2 - b2/4a2 + c/a = 0
    Rearrange to find x:
    (x + b/2a)2 = b2/4a2 - c/a
    x + b/2a = (+/-)sqrt[b2/4a2 - c/a]
    x = -b/2a (+/-) sqrt[b2/4a2 - c/a]
    Finally, fiddle around so that (1/2a) can be taken out as a common factor:
    x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]
    x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]
    x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)
    x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)
    x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.

  • Math -

    a bit easier to read ....

    divide by a

    x^2 + (b/a)x = -c/a
    complete the square , add b^2/(2a^2) to both sides

    x^2 + (b/a)x + b^2/(4a^2) = b^2/(4a^2) - c/a
    write the left side as a square and add the two terms on the right
    (x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2)
    take √ of both sides

    x + b/(2a) = ± √(b^2 - 4ac)/(2a)
    x = -b/(2a) ± √(b^2 - 4ac)/(2a)
    = (-b ± √(b^2 - 4ac)/(2a)

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