Question

Quadratic equations that can be solved using the quadratic formula can also be solved by factoring.

Question 1 options:
1) always
2) sometimes
3) never

Answer 1

sometimes

Answer 2

always
just not with integers

Answer 3

this question is stupid

Answer 4

i'll say. some math teacher problem

Answer 5

i mean you arent stupid for asking it marc
this is a stupid question to assign

Answer 6

Yeah, silly question. It can be factored if you don't use intergers, which is a somewhat difficult and lengthy process.

Answer 7

I really dont understand that last problem chaise..Solve 3x2 + 4x = 2.

Question 3 options:
1) 2 (+or-) √10 all over 6
2) -2 (+or-) √10 all over 3
3) -2 (+or-) √10 all over 3
4) -4 (+or-) √10 all over 3

can anyone help

Answer 8

right. factor theorem says if r is a root of a polynomial p(x) then p(x) can be factored as
\[p(x)-(x-r)q(x)\]

Answer 9

use quadratic formula after writing as
\[3x^2+4x-2=0\]

Answer 10

i.e. use
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=3,b=4,c=-2\]

Answer 11

get
\[x=\frac{-3\pm\sqrt{10}}{3}\] but it requires writing
\[\sqrt{16+24}=\sqrt{40}=\sqrt{4\times 10}=2\sqrt{10}\] and then canceling a 2 from top and bottom

Answer 12

that is why you get the answer i wrote.

Answer 13

The choice options are -2 and -4 \[\pm √10 / 3\]

Answer 14

\[x=\frac{-4 \pm \sqrt{4^2-4(3)(-2)}}{2(3)}=\frac{-4 \pm \sqrt{40}}{6}\]
\[=\frac{-4 \pm 2\sqrt{10}}{6}=\frac{-2 \pm \sqrt{10}}{3}\]

Answer 15

i think satellite meant to write a -2 not a -3 above
anyways the answer is what i have

Answer 16

Thats what i thought