Question

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Answer 1

Ok well first group the terms that have the same variable and move the constant to the other side of the problem

Answer 2

okay
idk how to do that

Answer 3

\[f(x) = 2x2 - 10x - 3\]

Answer 4

Thats the problem now do you know how to do what i just said?

Answer 5

dont i have to combine the x's?

Answer 6

Take out the f(x) = part because we know it would equal 0

Answer 7

so what would be the equation now?

Answer 8

can someone just give me the answer im freaking out here lol

Answer 9

or not argue about it

Answer 10

No because that's against the rules

Answer 11

u know what i mean

Answer 12

I can't just give you the answer's

Answer 13

That's why i'm going step by step wid you

Answer 14

ik i meant help

Answer 15

ok it would be 2x^2-3

Answer 16

No, when you take away the f(x) = part it would result in just "2x2 – 10x – 3 = 0"

Answer 17

ohh

Answer 18

Now just take away the "=0" and rewrite the problem to equal 3

Answer 19

\[2\times2-10x=\]3

Answer 20

Perfect, Now can you factor the coefficient?

Answer 21

idk how to do that

Answer 22

Well do yk what x would equal

Answer 23

no..

Answer 24

got it

Answer 25

can every comment before this be deleted please? thanks.

so a quadratic funtion is in the form of \( 0=a^2+bx +c \) right?

were given \( 0 = 2x^2 – 10x – 3 \)

so we can determine that:
\[ a=2 \]
\ \[c-3 \] now, we can either do it by facoring, or by using the quadratic formula -- factoring is kinda eh, so imma do it using the quadratic formula: \[ \frac{-b\pm\sqrt{b^2-4ac}}{2a} \Rightarrow \frac{10\pm\sqrt{-10^2-4 \times (2 \times -3 )}}{2 \times 2} \] now that we have that, we got some math to do: \[\frac{10\pm\sqrt{-10^2-4 \times -6}}{4}\] \[ \frac{10\pm\sqrt{100+24}}{4}\] \[ \frac{10\pm\sqrt{124}}{4}\Rightarrow\frac{10\pm\sqrt{\cancel{124}}}{\cancel{4}} \] \[ \frac{10\pm\sqrt{62}}{2} \Rightarrow \frac{\cancel{10}\pm\sqrt{\cancel{62}}}{2} \] \[\frac{5\pm\sqrt{31}}{2} \] now there are 2 possible solutions -- we will once do it the negative way, and the 2nd x will be the positive one -- it doesnt matter which one is firs, and which one is second -- before i do that, do YOU know how to do that?

Answer 26

@oliviac1111 ??

Answer 27

\[\frac{5\pm\sqrt{31}}{2} \]

\( \frac{5-\sqrt{31}}{2} \) AND \( \frac{5+\sqrt{31}}{2} \)

\[ \frac{5-\sqrt{31}}{2} \]
\[ \frac{5- 5.57 }{2} \]
\[ \frac{-0.57 }{2} \]
\[ x_1=-0.285 \]

lets solve for the second root.
\[ \frac{5+\sqrt{31}}{2} \]
\[ \frac{5+ 5.57 }{2} \]
\[ \frac{10.57 }{2} \]
\[ x_2=5.285 \]

and thats your answer.

Answer 28

thank u