Find Possible Values of k in Co-ordinate Geometry Question

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The discussion revolves around finding the possible values of k in a coordinate geometry problem involving points A, B, and C. The user initially attempted to use the Pythagorean theorem to derive equations for the distances AC and BC, leading to a quadratic equation that seemed incorrect. Another participant pointed out errors in the user's calculations and provided a corrected approach, ultimately arriving at a simplified quadratic equation. The correct equation, 3k² + 13k + 14 = 0, factors easily, suggesting that the user made a careless mistake in their initial work. The conversation emphasizes the importance of careful calculations in solving geometry problems.
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im stuck on an extension question to my hm/wk, hope you can help

Homework Statement



The points A, B and C have co-ordinate (-3, 2) (-1, -2), and (0, k) respectively, where k is constant.
Given that AC = 5BC, find the possible values of k.

Homework Equations





The Attempt at a Solution



I used pythagorus to get an equation for AC in terms of k, then the same for BC - then combined the 2 equations, simplified and ended up with 11 = 13k + 3k^2 then completed the square ending up with k = +-(squareroot)(301/36) - (13/6)

some how I think I've gone wrong or got the wrong method.

can someone please guide me :D thnx
 
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The method seems to be right. The answer is wrong. You'll need to show us the details of how you got the quadratic to solve.
 
My equation becomes \[<br /> 0 = 76 + 68k + 15k^2 <br /> \]<br />
 
... That suggests k = 2 or k= 2.53333333...repeating (two and eight-fifteenths)

..seems not to make sense.
 
The distance from A to C, squared, is (-3)2+ (2-k)2= 9+ 4- 4k+ k2. The distance from B to C, squared is (-1)2+ (-2-k)2= 1+ 4+ 4k+ k2. Saying that AC= 5BC is the same as AC2= 25BC2 or 13- 4k+ k2= 25(5+ 4k+ k2)= 125+ 100k+ 25k2. That gives 24k2+ 104k+ 112= 0 Dividing through by 8, that is 3k2+ 13k+ 14= 0. That doesn't look like what you got! And it factors rather easily! That's always a good sign.
 
thnx, i made a careless error, lol
 

Thread 'Greatest possible value of a constant in polynomial'

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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