Question

How do I use the quadratic formula to solve `3x^2 - 6 = 4x`?

Answer 1

`x = 2/3+-sqrt(22)/3`

Explanation:

Given:

`3x^2-6 = 4x`

Subtract `4x` from both sides to get:

`3x^2-4x-6 = 0`

This is now in standard form:

`ax^2+bx+c = 0`

with `a=3`, `b=-4` and `c=-6`

So we can apply the quadratic formula to find:

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

`color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(3))(color(blue)(-6))))/(2(color(blue)(3)))`

`color(white)(x) = (4+-sqrt(16+72))/6`

`color(white)(x) = (4+-sqrt(88))/6`

`color(white)(x) = (4+-sqrt(2^2 * 22))/6`

`color(white)(x) = (4+-2sqrt(22))/6`

`color(white)(x) = (2+-sqrt(22))/3`

`color(white)(x) = 2/3+-sqrt(22)/3`

Bonus

Notice in the derivation above that we had a factor `2^2` in the radicand that we brought outside the square root as `2` and then cancelled with the `2` in the divisor.

This always happens when `b` is even, adding an extra simplification step or two.

In such cases we can instead think of the original quadratic as taking the form:

`ax^2+2dx+c = 0`

Then the solutions can be expressed as:

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

In our example `a=3`, `d=-2` and `c=-6`, so we get:

`x = (-(color(blue)(-2))+-sqrt((color(blue)(-2))^2-(color(blue)(3))(color(blue)(-6))))/(color(blue)(3))`

`color(white)(x) = (2+-sqrt(4+18))/3`

`color(white)(x) = (2+-sqrt(22))/3`

`color(white)(x) = 2/3+-sqrt(22)/3`

I don't know how widely this variant of the quadratic formula is used.

If `a=1` then we can simplify further to say that the solutions of:

`x^2+2dx+c = 0`

are:

`x = -d+-sqrt(d^2-4c)`