Question

3x^2-3x-1>-1
please solve using quadratic inequalities

Answer 1

start with
\[3x^2-3x>0\] and factor as
\[3x(x-1)>0\] this one is zero at 0 and 1, and since it is a parabola that opens up it is positive outside the zeros, namely if \(x<0\) or if \(x>1\)

Answer 2

i really appriciate it. you have no idea:)

Answer 3

it is clear (more or less) how to do it? you have to have a zero on one side of the inequality
then you need to find the actual zeros
then think about whether you want positive or negative

Answer 4

we can work through another one if you have one more to do

Answer 5

its clear! i just want to work one more because the sign > has a line underneath it

Answer 6

-x^2+x+2>0

Answer 7

then so will your answer

Answer 8

i hate working with negative leading coefficients, so if this is
\[-x^2+x+2\geq 0\] then i would change the sign and the inequality and start with
\[x^2-x-2\leq 0\]

Answer 9

again we factor, this time as
\[(x+1)(x-2)\] so the zeros are at \(x=-1\) and at \(x=2\)

Answer 10

we are looking for where it is less than or equal to zero, and again this is a parabola that faces up, so it is negative between the zeros i.e. if
\[-1\leq x\leq 2\]

Answer 11

it is much more convenient to work with a positive leading coefficient, rather than a negative one
that is why i changed
\[-x^2+x+2\geq 0\] to
\[x^2-x-2\leq 0\]

Answer 12

wow i see that makes sense!!

Answer 13

i am so happy that i found this site

Answer 14

glad to help

Answer 15

you are a great person for helping others :) thank you this really makes sense!

Answer 16

yw

Answer 17

if you dont mind helping me with one more?

Answer 18

4x^2+4x>3

Answer 19

the > has a line underneath it also