solution method ofquadratic equations

[petr]

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]

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
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]

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.

 

[Deleted member 4993]

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]

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?

 

[Deleted member 4993]

but where the 5y( is lost?

There are two solutions.

The other solution is y = 0

 

[petr]

thank you

There are two solutions.

The other solution is y = 0

ohh i ve just got it

thank you.