Hi,

I know how to solve a quadratic equation but I have a book where nearly all quadratics problems are finished in this form>

for (4y^2-20y+25) +y^2 = 25

----------------

> 5y(y-4) = 0 <---- this form

----------------

what means that y1 = 4 and y2 = 0

why can they just take the numbers out of this form? and how is this form called? where can i get some more information about it?

Thank you,

Peter

A polynomial, in particular a quadratic, can be written as a constant times the product of x-x

_{i} where the x

_{i} are its roots. For example

x

^{2}-4 = (x-2)(x+2)

To see this in general, take any quadratic

y = a x

^{2} + b x + c

where a is not zero. We know (from the quadratic formula) that the zeros (roots) of the quadratic equation can be written as

\(\displaystyle x_1\, =\, \frac{-b\, +\, d}{2\, a}\)

and

\(\displaystyle x_2\, =\, \frac{-b\, -\, d}{2\, a}\)

where

\(\displaystyle d\, =\, \sqrt{b^2\, -\, 4\, a\, c}\)

Thus

y = A (x-x

_{1}) (x-x

_{2})

where A is some constant. If you multiple this out we get

y = A (x

^{2} -(x

_{1}+x

_{2}) x + x

_{1} x

_{2})

= A ( \(\displaystyle x^2\, +\, 2\, \frac{b}{2\, a}\, x\, +\, \frac{b^2\, -\,d^2}{4\, a^2}\) )

= A ( \(\displaystyle x^2\, +\, \frac{b}{a}\, x\, +\, \frac{4\, a\, c}{4\, a^2}\) )

= A ( \(\displaystyle x^2\, +\, \frac{b}{a}\, x\, +\, \frac{c}{a}\) )

= \(\displaystyle A\, x^2\, +\, (\frac{A}{a})\, b\, x\, +\, (\frac{A}{a})\, c\)

So we see that, if we let A=a, we have our original equation.

One reason why one might want to do this is that the zeros are readily identified.