Question

Use the quadratic formula to find the zeros of the function. Round to tenths if necessary. y = 3x2 + 5x
A. {-2.23, 1.7}
B. {-1.7, 0}
C. {-2.8, 1.7}
D. no real solution

Answer 1

a = 3
b = 5
c = 0

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \frac{-5 \pm \sqrt{5^2 - 4(3)(0)}}{2(3)}\]
\[x = \frac{-5 \pm \sqrt{5^2}}{2(3)}\]

Answer 2

You may be able to finish solving from there.

Answer 3

so no real solution?

Answer 4

Did you forget the c value when you were writing the quadratic equation?

Answer 5

The solution according to what you posted is (0,0)

Answer 6

so 0,-1.7

Answer 7

Which is not in the answer choices...

Answer 8

No, that is not a solution

Answer 9

y = 3(0)^2 + 5(0)
= 0 + 0
= 0

Answer 10

(0,0) works
(0,-1.7) does not work

Answer 11

(-1.7,0) doesn't work either

Answer 12

In fact, none of the given answer choices work

Answer 13

I'm pretty certain you wrote the problem incorrectly to begin with because the answer choices do not match the question.

Answer 14

\[-5\pm \sqrt{25-0}/6 \] thats the equation i get

Answer 15

Once again, the answer choices do not match the problem. Double check to make sure you have posted it from your book correctly.

Answer 16

y = 3x2 + 5x

Answer 17

the 2 is a exponent

Answer 18

You're telling me something I already know.

Answer 19

You insert the coefficients a, b, and c into the quadratic formula.

Answer 20

But you don't seem to have a c value. A quadratic equation in general form is
\(ax^2 + bx + c = 0\)

Answer 21

You have \(3x^2 + 5x = 0\) but you don't have a \(c\) value.

Answer 22

the question asks to fine the 0

Answer 23

find

Answer 24

I found the zero.

Answer 25

Use the quadratic formula to find the zeros of the function. Round to tenths if necessary

Answer 26

x = 0 and y = 0

I already demonstrated it to you above.

Answer 27

There is something about what I did that you don't understand apparently.

Answer 28

i get it. but your not giving a answer. so i dont get what your talking about now