Quadratic Equation Solving

manasan

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Jul 21, 2019
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\(\displaystyle if, \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \\ Show \ that, \frac{ 2a^4b^2 + 3a^2c^2-5e^4f }{ 2b^6 + 3b^2d^2-5f^5 } = \frac{a^2ce}{b^2df} \)
I tried making a,b,c as subject but it resulted in two variable equation which was unable to be plugged into the 2nd equation and arrive in single variable solution.
So I don't even know where to start actually?
 

Dr.Peterson

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I would set \(\displaystyle \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k\), and replace a, c, and e with expressions in terms of b, d, and f. See what happens.
 

manasan

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Thank You I solved, but is there any other short way to solve.
\(\displaystyle By\ substituting\ {\text{'k'}}\ I\ got\ \frac{2 k^{2} a^2 b^{4}+3 k^{4} b^{2} d^{2}-5 k^{3} f^{4} e}{2 a^{6} k^{-6}+3 k^{4} a^{2} c^{2}-5 e^{5} k^{-5}}=k^4\\ \text{and I divided by}\ k^4\ \text{and solving lead me to same denominator and numerator on one side and 1 on the other side.}\)
Is this the shortest route or are there any better methods?
What should I do to tackle these kinds of problems easily and quickly next time?😅
 

Dr.Peterson

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This is basically a standard technique when you are given a proportion, especially one like this one.

It doesn't look like you did quite what I suggested, so maybe your work was harder than mine. I replaced a with bk, c with ck, and e with fk everywhere on both sides, and both sides easily become \(\displaystyle k^4\). I'm also not sure how you proceeded from what you show.

Try doing what I suggested, and show me your work if it isn't really easy, so I can make suggestions.
 
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