Question

how do to solve quadratic inequalities ??

Answer 1

The same way you do quadratic equations for the most part. The only difference is that at the end you'll end up with ranges rather than distinct values.

Answer 2

actually the problem is with the sign of range ..i mean which direction ??

Answer 3

find the zeros. if there are none then the quadratic is always positive if the leading coefficient is, negative if leading coefficient is negative

Answer 4

if the quadratic has zeros as
\[r_1,r_2\] with
\[r_1<r_2\] then look at leading coefficient
if it is positive your quadratic will be positive of on
\[(\infty,r_1)\cup (r_2,\infty)\]

Answer 5

Yeah, it's easier to show you if you have an example.

Answer 6

sorry that should be
\[(-\infty,r_1)\cup (r_2,\infty)\]

Answer 7

what do mean by zeros ?

Answer 8

and if the leading coefficient is negative then you quadratic will be positive between the zeros, on
\[(r_1,r_2)\]

Answer 9

The values for x that cause the expression to simplify to 0.

Answer 10

i means the values for which your quadratic is 0. lets do an example

Answer 11

\[x^2-5x+6>0\]
first factor to find the zeros get
\[x^2-5x+6=0\]
\[(x-2)(x-3)=0\]
\[x=2\] or \[x=3\] so zeros are 2 and 3

Answer 12

now
\[y=x^2-5x+6\] is a parabola that faces up because the leading coefficient it positive (it is 1). so this thing will be positive on the interval
\[(-\infty,2)\cup(3,\infty)\] and negative on
\[(2,3)\]

Answer 13

and don't forget that when you solve an inequality you have done more than just solved the specific one asked for. so if you were asked
\[x^2-5x+6>0\] answer is
\[(-\infty,2)\cup(3,\infty)\] but you also know that the solution to
\[x^2-5x+6<0\] is
\[(2,3)\]