Finding zeros (math homework)

[Pat Fairbank]

Originally Posted by VEN

So say my relation is A = w(20-w), I'd solve that by making w=0 and 20-w = 0? So all this is, is substitution? Guess and check?

Let me explain the whole derivation of the quadratic formula.

You have a quadratic equation of the form y = ax^2 + bx + c. You want to find the roots (zeros) of the equation, meaning the values of x at which y is 0. So the equation takes the form ax^2 + bx + c = 0. (Or if the equation already has this form, you can start from here.)

Now, all we have to do is solve for x.

First, you factor the a out of ax^2 + bx, and put the c on the other side:

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

Next, add b^2/(4a^2) inside the square brackets to complete the square (gotten by dividing b/a by 2 and squaring). Multiply this value by a and add it to the other side to equalize things:

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

The contents of the square brackets can now be reduced to [x + b/(2a)]^2, and at the same time we'll make the right side into one fraction:

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

Divide both sides by a:

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

Take the square root of both sides:

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

Isolate x:

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

And now you have a formula that can solve for x whenever you have an equation that looks like 0 = ax^2 + bx + c.

[VEN]

grr.....why does it have to be so confusing.....at least to me

Thanks though. I'll try and understand