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If a, b, c is in G.P, show that equation $ a{x^2} + 2bx + c = 0 $ and $ d{x^2} + 2ex + f = 0 $ have a common root provided that $ \dfrac{d}{e},\dfrac{e}{b},\dfrac{f}{c} $ are in A.P.

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Last updated date: 23rd Jul 2024
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Answer
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Hint: First we have to define what the terms we need to solve the problem are.
The given question they were asking to find the common roots, solution as follows, since we need to know about Arithmetic progression. An arithmetic progression can be given by $ a,(a + d),(a + 2d),(a + 3d),... $ where $ a $ is the first term and $ d $ is the common difference.

Complete step by step answer:
Since in this question we need prove $ \dfrac{d}{e},\dfrac{e}{b},\dfrac{f}{c} $ are in the common roots of the quadratic terms; $ a{x^2} + 2bx + c = 0 $ and $ d{x^2} + 2ex + f = 0 $ in the arithmetic progression.
Let the question is in the form of quadratic equation take the first equation, $ a{x^2} + 2bx + c = 0 $
In this equation assume that for the quadratic equation general formula $ {b^2} = ac $ or $ b = \sqrt {ac} $ and we will substitute in the equation which we take above we get; $ a{x^2} + 2bx + c = 0 $ $ \Rightarrow a{x^2} + 2\sqrt {ac} x + c = 0 $ (now we are going to take the common terms out which like $ {(a + b)^2} = {a^2} + {b^2} + 2ab $ ) we get; $ a{x^2} + 2\sqrt {ac} x + c = 0 \Rightarrow {(x\sqrt a + \sqrt c )^2} = 0 $ and the square will go to the right side and cancelled we get $ x = - \dfrac{{\sqrt c }}{{\sqrt a }} $ (the equator will term to right side) is the value of x;
Now we have the value of $ x = - \dfrac{{\sqrt c }}{{\sqrt a }} $ and substitute in the second equation $ d{x^2} + 2ex + f = 0 $
Thus $ d(\dfrac{c}{a}) + 2e( - \dfrac{{\sqrt c }}{{\sqrt a }}) + f = 0 $ (now taking the common terms and cross multiplying we get) after further solving into the simplified form we get $ \dfrac{d}{a} + \dfrac{f}{c} = 2e\sqrt {\dfrac{1}{{ac}}} $ (cross multiplied and changes of right left side equation also the common roots taken out)
since $ {b^2} = ac $ thus we get $ \dfrac{d}{a} + \dfrac{f}{c} = \dfrac{{2e}}{b} $ (which is in the addition with respect to A.P)
Thus, the quadratic equations of the first equation $ a{x^2} + 2bx + c = 0 $ and the second equation $ d{x^2} + 2ex + f = 0 $ have a common root provided that $ \dfrac{d}{e},\dfrac{e}{b},\dfrac{f}{c} $ are in A.P.

Note: likewise in G.P the terms are, \[a,(ar),(a{r^2}),(a{r^3}),...\] where $ a $ is the first term and $ r $ is the common ratio.
The quadratic equation can be simplified $ a{x^2} + 2bx + c = 0 $ into $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ so that we can able to find the roots of the quadratic equation.