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$$\text{Let p(x)=\sqrt{-x}, and q(x)=8x^2+10x-3. The domain of p(q(x)) can be written in the form a\le x \le b. Find b-a. }$$

Jul 18, 2019

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p (q(x))  =  √ [ - ( 8x^2 + 10x - 3) ]

For the domain to be defined, - (8x^2 + 10x - 3 )  must be ≥ 0

So

-(8x^2 + 10x - 3)  ≥ 0

-8x^2 - 10x + 3 ≥ 0

To solve this, set this = 0    and solve for x

-8x^2 - 10x + 3  = 0          multiply through by -1

8x^2 + 10x - 3  = 0           factor

(4x - 1) ( 2x + 3)   = 0

Set each factor to  0  and solve for x

4x - 1  = 0           and        2x + 3  = 0

4x  = 1                               2x   = - 3

x = 1/4                                 x  = - 3/2

The solution will either come from  x  = (-inf, -3/2] U [1/4, infinity)  or   x  = [-3/2, 1/4]

Testing a point in the second interval  [ I'll pick x = 0 ].....we can see that

√ [ - ( 8(0)^2 + 10(0) - 3) ]   =   √ [ - ( -3) ]  =  √ 3     gives us a real number

So....the interval that solves this is   [ -3/2, 1/4 ]

So  b  - a  =    1/4  - (-3/2)  =  (1/4) - (-6/4)  =  7/4

Here's a graph to  show that the solution interval is correct :

https://www.desmos.com/calculator/6jhmgavfr0   Jul 18, 2019