Find the length of PA - deductive Geometry

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The discussion focuses on solving for the length of segment PA using the equation BP × AP = PT^2. By setting AP as x, the equation simplifies to (6+x)x=16, leading to the quadratic x^2 + 6x - 16 = 0. The solutions yield x=2 as the only valid positive value for PA. The theorem used in the solution is referenced for further proof in a provided link. The approach is confirmed as correct by participants in the discussion.
chwala
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Homework Statement
see attached
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Geometry
Text question is here and solution;

1665266504194.png

My approach;
##BP ×AP= PT^2##
Let ##AP= x##
Therefore, ##(6+x)x=16##
##x^2+6x-16=0##
##x=2## or##x=-8##
##⇒x=2## positive value only.

I guess this may be the only approach. Cheers!
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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