Question

find the zeros of the equation.

y=3x^2 -21x+3
If somebody could get me started i can get the rest :)

Answer 1

Factor out 3

Answer 2

y=3x(x-7)+3?

Answer 3

Did you read where I said to factor out the 3?

Answer 4

I'm assuming I did that wrong.

Answer 5

3 is different from 3x

Answer 6

Therefore, factoring out 3 is different from factoring out 3x

Answer 7

\[3x^2-21x+3=3(x^2-7x+1)\]

Answer 8

3(x^2-7x+1)
Looks like you're going to need the good ol' quadratic formula.

Answer 9

i got 7+/-3square root of 5
all over 2

Answer 10

When we talk about the "zeros" of a function, we will reffer to the "roots" of such.
what are the roots of a function? we call roots to all those values of x, that gives us a image of 0.
So, what you are showing us is a 2nd degree equation, wich we can also state as Ax^2+Bx+C.
Now to find the roots of that equation, we will have to find an equation that allows us to find it's roots, with that, I mean the values of "x" that makes it equal 0.
\[Ax ^{2}+Bx+C=0\]
let 2m=B/A and n=C/A
then:
\[x ^{2}+2mx+n=0\]
I'll add m^2 to both sides and sustract n:
\[x ^{2}+2mx+m ^{2}=m ^{2}-n\]
Let's simplify:
\[(x+m)^{2}=m ^{2}-n\]
\[x+m=\pm \sqrt{m ^{2}-n}\]
sustracting m to both sides:
\[x=-m \pm \sqrt{m ^{2}-n}\]
now, let's replace all thos variables to their original form:
\[x=-\frac{ B }{ 2A }\pm \sqrt{(\frac{ B }{ 2A })^{2}-\frac{ C }{ A }}\]
now it's just a matter of a fractionary addition:
\[x=\frac{ -B \pm \sqrt{B ^{2}-4AC} }{ 2A }\]
Would you look at that, I've proven to you, that not only the values of the roots are calculated in thar formula, only applyable on 2nd degree equations, but that there are two of them.

Let's take a look at the problem:
\[y=3x ^{2}-21x+3\]
it has the form Ax^2+Bx+C, so to find it's root, we will make it equal 0 and apply the formula I've just deduced, but before that, let's take out 3 as common factor:
\[y=3(x ^{2}-7x+1) \]
\[y=0\]
a property of Hankel states that if ab=0 then either b=0 or a=0.
so, we will say:
\[3=0\]
or
\[x ^{2}-7x+1=0\]
wich is evident by now, because 3 is not equal 0.
so let's apply the formula I deduced but I'll teach you a trick I used to not confuse the terms:
\[x ^{2}+(-7)x+1=0\]
so we can say: A=1, B=-7 and C=1
so the roots will be:
\[x=\frac{ -(-7)\pm \sqrt{(-7)^{2}-4(1)(1)} }{ 2(1) }\]
\[x=\frac{ 7\pm \sqrt{45} }{ 2 }\]
so the roots, let's call them x1 and x2, will be:
\[x _{1}=\frac{ 7+\sqrt{45} }{ 2 }\]
\[x _{2}=\frac{ 7-\sqrt{45} }{ 2 }\]