Question

Solve using algebra. 6x^2-13x+6≥0

Answer 1

\[(3x-2)(2x-3)\ge\]
\[x \ge2/3, x \ge3/2\]

Answer 2

start with
\[(2x-3)(3x-2)\geq 0\] then see that the zeros are at
\[\frac{2}{3}\] and
\[\frac{3}{2}\] since you have a quadrant with positive leading coefficient, it is a parabola that opens up so it will be positive outside the roots and negative between them. you want positive, so answer is
\[(-\infty, \frac{2}{3}]\cup (\frac{3}{2},\infty)\]

Answer 3

*quadratic, not "quadrant"

Answer 4

i dont get why it is negative infinity to 2/3???

Answer 5

because it is every number except those that we found

Answer 6

\[y=6x^2-13x+6\] looks like this

Answer 7

it is negative (below the x axis) between the zeros and positive (above the x axis) outside the zeros

Answer 8

ya but if not looking at a graph, how would you know?

Answer 9

you don't actually have to draw the picture, that is just my explanation. a quadratic with positive leading coefficient will be negative between the zeros and positive elsewhere

Answer 10

all quadratics look the same. you will not get a different picture. it will either open facing up or facing down depending on the leading coefficient

Answer 11

ohhhh ok i get it now i see. thanks satellite :)

Answer 12

yw