how do you find the vertex of 25x^2-30x+9=0 ???

Question

So Vertex: -b/2(a)
when i plug that in i get 30/2(25) = 30/50= 3/5

But the answer is -4 why cant i get it?!

Answer 1

For one thing, I think you want to find the vertex of

y = 25x^2 - 30x + 9

because equations don't have vertices.

y = 25x^2 - 30x + 9

First, factor out 25.

y = 25(x^2 - (30/25)x) + 9

Now, reduce 30/25.

y = 25(x^2 - (6/5)x) + 9

Take "half squared" of -6/5, add it in the brackets, and offset 25 times that value outside the brackets.

y = 25(x^2 - (6/5)x + 9/25) + 9 - 9

y = 25(x - (3/5))^2 + 0

Therefore, the vertex is at (3/5, 0).

I got the same answer as you.

Answer 2

x={ b-/+ radical(b^2 -4ac)} /2a

also the answer is -0.6=(-3/5)
and not -4
u can try it, put -4 to see it:
25(16) - 30(-4)+9 =529 and that's not 0 !!
but
25(-0.6) - 30(-0.6) +9= 9-18+9=0 and that's right!

Answer 3

you should do this work :
b^2-4ac=z
x1=(b-(z^.5))/2
x2=(b+(z^.5))/2