solution method ofquadratic equations

petr

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Oct 22, 2015
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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

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Joined
Jul 30, 2014
Messages
3,345
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
Joined
Oct 22, 2015
Messages
24
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

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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

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Subhotosh Khan

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petr

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