Question

How do you solve `x^2+4x-3=0` using the quadratic formula?

Answer 1

Use the quadratic formula to find:

`x = -2+-sqrt(7)`

Explanation:

`x^2+4x-3=0` is of the form `ax^2+bx+c=0`, with `a=1`, `b=4` and `c=-3`.

This has roots given by the quadratic formula:

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

`=(-4+-sqrt(4^2-(4*1*(-3))))/(2*1)`

`=(-4+-sqrt(16+12))/2`

`=(-4+-sqrt(28))/2`

`=(-4+-sqrt(2^2*7))/2`

`=(-4+-2sqrt(7))/2`

`=-2+-sqrt(7)`

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Using the quadratic formula is similar in some ways to completing the square:

`0 = x^2+4x-3`

`=x^2+4x+4-7`

`=(x+2)^2-(sqrt(7))^2`

`=((x+2)-sqrt(7))((x+2)+sqrt(7))`

`=(x+2-sqrt(7))(x+2+sqrt(7))`

Hence `x = -2+-sqrt(7)`

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In fact, we can derive the quadratic formula as follows:

Given `ax^2+bx+c = 0`

`0 = ax^2+bx+c`

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

Hence:

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

Divide both sides by `a` to get:

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

Hence:

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

So:

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