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# Help! Fast!

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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 4, 2018

#1
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p( q(x) )   =   √[ -(8x2 + 10x - 3) ]

To avoid the square root of a neative number, the domain must be the  x  values such that

-( 8x2 + 10x - 3)   ≥   0

8x2 + 10x - 3   ≤   0

To solve this inequality, let's first find the  x  values that make

8x2 + 10x - 3   =   0

8x2 + 12x - 2x - 3   =   0

4x(2x + 3) - 1(2x + 3)   =   0

(2x + 3)(4x - 1)   =   0

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

x   =  -3/2             or            x  =  1/4

Since  8x2 + 10x - 3  is a quadratic equation, we can say that the  x  values that make it ≤ 0 will be either

x ≤ -3/2   and   x ≥ 1/4           OR          -3/2 ≤ x ≤ 1/4

If  x = -2 ,     8x2 + 10x - 3  =  8(-2)2 + 10(-2) - 3  =  24 - 20 - 3  =  1

So the  x  values in the interval   x ≤ -3/2  and  x ≥ 1/4   do not make   8x2 + 10x - 3  ≤  0

If  x = 0 ,     8x2 + 10x - 3  =  8(0)2 + 10(0) - 3  =  -3

So the  x  values in the interval   -3/2 ≤ x ≤ 1/4   do make   8x2 + 10x - 3  ≤  0

The domain of  p( q(x) )   is   -3/2 ≤ x ≤ 1/4

1/4 - -3/2   =   1/4 + 3/2   =   1/4 + 6/4   =   7/4

Feb 4, 2018
#2
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Very Nice solution, hectictar! Great! Feb 5, 2018