Question

what are the zeros of h(x)=2x^2-16x+27

Answer 1

As it's a quadratic equation you can simply use this formula: \[
x_{1,2} = \frac{1}{2a}\left[-b\pm \sqrt{b^2-4ac}\right]
\]

Answer 2

Do you know it?

Answer 3

to find the possible zeros?

Answer 4

yes

Answer 5

it's basically the standard formula

Answer 6

but i think there is more then 1 zero

Answer 7

okay let me show you

Answer 8

thank you because i dont understand

Answer 9

??

Answer 10

\[h(x) = \underbrace{2}_{a}x^2-\underbrace{16}_{b}x+\underbrace{27}_c\]
Now we plug that into the formula:\[
x_{1/2} = \frac{1}{4}\left[ -(-16) \pm \sqrt{16^2 - 4\cdot2\cdot27}\right]
\]
\[
= \frac{1}{4}\left[ 16 \pm \sqrt{256 - 216}\right]
\]
\[
= \frac{1}{4}\left[ 16 \pm \sqrt{40}\right]
\]
\[
= \frac{1}{2}\left[ 8 \pm 2\sqrt{10}\right]
\]
Now we can get the zeros as:\[
x_1 = \frac{1}{2}\left[ 8 + 2\sqrt{10}\right], \quad x_2 = \frac{1}{2}\left[ 8 - 2\sqrt{10}\right]
\]

Answer 11

i got the same answer, thanks so much

Answer 12

You're right that there are two solutions, they exists because we take the squareroot of something. Which means there are necessarily two solutions, because the result of that squareroot can either be negative or positive. \[
\sqrt{-a^2} = \sqrt{-a(-a)} = \sqrt{a^2} = a
\]
So we cannot tell whether which means we cannot tell whether a was positive or negative int the beginning.

Answer 13

You're welcome

Answer 14

how do i find the y- intercept of f(x)= (x^2-6x-7)/(4x^2-36x+56)