Answer:
5. (5x^7)√(5x)
6. x = -2
Step-by-step explanation:
5.
[tex]\displaystyle\frac{\sqrt{250x^{16}}}{\sqrt{2x}}=\sqrt{\frac{250x^{16}}{2x}}=\sqrt{125x^{15}}\\\\=\sqrt{(25x^{14})(5x)}=5x^7\sqrt{5x}[/tex]
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6.
[tex]\sqrt{2x+13}-5=x\\\\\sqrt{2x+13}=x+5 \qquad\text{add 5}\\\\2x+13=x^2+10x+25 \qquad\text{square both sides}\\\\x^2+8x+12=0 \qquad\text{subtract the left side}\\\\(x+6)(x+2)=0 \qquad\text{factor}\\\\x=-6 \quad\text{or}\quad x=-2[/tex]
There is always the possibility of extraneous solutions for equations like this. We can check.
√(2·(-6)+13) -5 = -6 . . . . substitute -6 for x
√1 -5 = -6 . . . . . . . . . . . . not true; -6 is an extraneous solution
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√2·(-2)+13) -5 = -2
√9 -5 = -2 . . . . . . . . . . . true; x = -2 is the solution
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The attached graph has the equation rewritten so it is of the form
f(x) = 0
where
f(x) = √(2x+13) -5 -x
The x-intercept is highlighted on the graph. It is x=-2. You can see that if the negative branch of the square root function were included, it might make an x-intercept at x=-6. For our purposes, the square root function is the positive square root only, so that branch is not included and -6 is an extraneous solution.
