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)+(b)/(a)x +(c)/(a) =0`

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

`x^(2) +(b)/(a)x =-(c)/(a)`

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

`(x+(b)/(2a))^(2) - (b^(2) )/(4a^(2)) =-(c)/(a)` (now add the constant in terms of a and b)

`(x+(b)/(2a))^(2) = -(c)/(a) +(b^(2))/(4a^(2))` (now simplify the fractions on the right side)

`(x+(b)/(2a))^(2)= (-c(4a)+b^(2))/(4a^(2))`

`(x+(b)/(2a))^(2)= (-4ac+b^(2))/(4a^(2))` (now just put the b^2 in front)

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

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

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

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

5. Now just solve for x:

`x= (√(b^2-4ac))/(2a)-(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.