Question

Derive the quadratic formula from the standard form (ax² + bx + c = 0) of a quadratic equation by
following the steps below.
1. Divide all terms in the equation by a.
2. Subtract the constant (the term without an x) from both sides.
3. Add a constant (in terms of a and b) that will complete the square.
4. Take the square root of both sides of the equation.
5. Solve for x.​

Answer 1

1. Divide all terms by a:

$x^{2}+\frac{b}{a}x +\frac{c}{a} =0$

2. Subtract the constant (which is c/a):

$x^{2} +\frac{b}{a}x =-\frac{c}{a}$

3. Complete the square, then add the constant to both sides:

$(x+\frac{b}{2a})^{2} - \frac{b^{2} }{4a^{2}} =-\frac{c}{a}$ (now add the constant in terms of a and b)

$(x+\frac{b}{2a})^{2} = -\frac{c}{a} +\frac{b^{2}}{4a^{2}}$ (now simplify the fractions on the right side)

$(x+\frac{b}{2a})^{2}= \frac{-c(4a)+b^{2}}{4a^{2}}$

$(x+\frac{b}{2a})^{2}= \frac{-4ac+b^{2}}{4a^{2}}$ (now just put the b^2 in front)

$(x+\frac{b}{2a})^{2}= \frac{b^{2}-4ac}{4a^{2}}$

4. Square root both sides and simplify the right side:

$x+\frac{b}{2a} = \sqrt{\frac{b^{2}-4ac}{4a^{2}} }$ (you can square root the bottom bit of the fraction fully)

$x+\frac{b}{2a} = \frac{\sqrt{b^{2}-4ac} }{2a}$

5. Now just solve for x:

$x= \frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$ (now simplify)

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ (note: it should be a plus or minus sign infront of the squareroot, not just a plus sign -it's just that i can't write it in )

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If you have any questions, feel free to ask.