finding domain of a square root

[smarteepantz]

Domain:

H(x)= sqrt(2x^2 + x -6)

i understand it cannot be negative under a square root. i don't know how to solve this

 

[DanieldeLucena]

It will be a inequation 2x^2 + x - 6 >0

Can you solve this ????

 

[smarteepantz]

(2x-3)(x+2) >= 0 ?

i did try that, but not sure where to go from here?

 

[DanieldeLucena]

Okay you're right >=
So you have the roots -2 and 3/2 like it
[attachment=0:tdw17e1w]1.JPG[/attachment:tdw17e1w]

 

[DanieldeLucena]

you want y >= 0
[attachment=0:3re2z6je]1.JPG[/attachment:3re2z6je]
So you'll see too that y >= 0 when the roots is x <= -2 or when x >= 3/2

okay now ???!!!!

 

[smarteepantz]

how did you get <= -2

 

[lookagain]

Okay you're right >=
So you have the roots -2 and 3/2 like it
[attachment=0:6js4ksd3]1.JPG[/attachment:6js4ksd3]

DanieldeLucena,

the different values of x are to have subscripts. As you have them,
they are exponents. And the quadratic can be greater than or equal
to zero.

The discriminant, as you indicated by that type of triangle symbol, is

\(\displaystyle b^2 - 4ac =\)

\(\displaystyle (1)^2 - 4(2)(-6) =\)

\(\displaystyle 49\)



\(\displaystyle 2x^2 + x - 6 \ge 0\)


\(\displaystyle x_1 = \frac{1 - \sqrt{49}}{2(2)}\)

\(\displaystyle x_1 = \frac{- 1 - 7}{4}\)

\(\displaystyle x_1 = \frac{-8}{4}\)

\(\displaystyle x_1 = -2\)

------------------------------------------------------------------

\(\displaystyle x_2 = \frac{-1 + \sqrt{49}}{2(2)}\)

\(\displaystyle x_2 = \frac{-1 + 7}{4}\)

\(\displaystyle x_2 = \frac{6}{4}\)

\(\displaystyle x_2 = \frac{3}{2}\)

-----------------------------------------------------------------

Then the critical numbers to use for the inequality are

\(\displaystyle x_1 = -2, \ x_2 = \frac{3}{2}\)