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# need help asap

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Let $$p(x)=\sqrt{-x}$$, and $$q(x)=8x^2+10x-3$$. What is the domain of $$p(q(x))$$? Your answer will be of the form $$a\le x \le b$$. Find $$b-a$$.

Feb 10, 2018

### 1+0 Answers

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

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

We want to find the x values that make

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

8x^2  + 10x  -  3  =   0  factor

(4x  - 1) (2x + 3)  = 0

Setting both factors to 0 and solving for x we have that  x  = 1/4   and x  = -3/2

However......the quantity under the radical, 3 - 8x^2 - 10x   must be ≥ 0

So...we have three possible intervals that may solve this

(-inf, -3/2] U [ -3/2, 1/4}] U {1/4, inf )

We only need to test a point in  the middle interval..if it "works," this interval will solve the equation

3 - 8x^2 - 10x ≥  0

Testing  x  = 0  will make this true, so the interval that solves the equation is

-3/2  ≤ x  ≤ 1/4

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

Feb 10, 2018