Question

If ${b^2} = ac$ equation $a{x^2} + 2bx + c = 0$ and $d{x^2} + 2ex + f = 0$ have a common root then $\dfrac{d}{e},\dfrac{e}{b},\dfrac{f}{c}$ are in
$1)A.P$
$2)G.P$
$3)H.P$
$4)$ None of these

Answer 1

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 and in which progression it will occur, solution as follows
since we need to know about Arithmetic progression. An arithmetic progression can be expressed by $a,(a + d),(a + 2d),(a + 3d),...$ where $a$ is the first term and $d$ is a common difference.

Complete step-by-step solution:
Since in this question we need to 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 $d{x^2} + 2ex + f = 0$ are the given equations,
Let the question is in the form of a 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 canceled we get $x = - \dfrac{{\sqrt c }}{{\sqrt a }}$ (the equator will term to the 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
Hence, the option $1)A.P$ is correct.

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.