# solution method ofquadratic equations

#### petr

##### New member
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

#### Ishuda

##### Elite Member
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-xi where the xi are its roots. For example
x2-4 = (x-2)(x+2)
To see this in general, take any quadratic
y = a x2 + 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-x1) (x-x2)
where A is some constant. If you multiple this out we get
y = A (x2 -(x1+x2) x + x1 x2)
= 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.

#### petr

##### New member
y equal to something

What are you asking?
Simplifying:
4y^2 - 20y + 25 + y^2 - 25 = 0
5y^2 - 20y = 0
y^2 - 4y = 0
ky(y - 4) = 0 : k can be any value...
I don't understand why it is possible to take the value just out of the parentheses and write that y=4 while out of the brackets are still some variables.

If i want to take the value of y I need to count it all, don't? Expand the brackets and make y equal to some particular value.

but in this context it is just taken out of the equation y-4 = 0 => y = 4.

#### Subhotosh Khan

##### Super Moderator
Staff member
I don't understand why it is possible to take the value just out of the parentheses and write that y=4 while out of the brackets are still some variables.

If i want to take the value of y I need to count it all, don't? Expand the brackets and make y equal to some particular value.

but in this context it is just taken out of the equation y-4 = 0 => y = 4.
If

A * B = 0

Then

either

A = 0

or

B= 0

So when

y(y-4) = 0

then

either

y - 4 = 0 → y = 4

or

y = 0

#### petr

##### New member
of course

If

A * B = 0

Then

either

A = 0

or

B= 0

So when

y(y-4) = 0

then

either

y - 4 = 0 → y = 4

or

y = 0
but where the 5y( is lost?

#### Subhotosh Khan

##### Super Moderator
Staff member
but where the 5y( is lost?
There are two solutions.

The other solution is y = 0

#### petr

##### New member
thank you

There are two solutions.

The other solution is y = 0

ohh i ve just got it thank you.