Question

If three distinct numbers a, b, c are in G.P and the equations $a{x}^2+2bx+c=0$ and$d{x}^2+2ex+f=0$ have a common root, then which one of the following statements is correct?
(a) d, e, f are in A.P.
(b) ${\displaystyle \frac{d}{a}}$, ${\displaystyle \frac{e}{b}}$, ${\displaystyle \frac{f}{c}}$ are in G.P.
(c) ${\displaystyle \frac{d}{a}}$, ${\displaystyle \frac{e}{b}}$, ${\displaystyle \frac{f}{c}}$ are in A.P.
(a) d, e, f are in G.P.

Answer 1

Hint:We start solving the problem by recalling the fact that if three numbers p, q, and r are in G.P (Geometric Progression), then $q^2=pr$. We use this fact and make a substitution in the quadratic equation $a{x}^2+2bx+c=0$ to find the roots of the equation. We then substitute the obtained root in the quadratic equation $d{x}^2+2ex+f=0$ and make necessary calculations and arrangements to get the desired relation.

Complete step by step answer:
According to the problem, we have three distinct numbers a, b, c which are in G.P. We have a common root for the quadratic equations $a{x}^2+2bx+c=0$ and$d{x}^2+2ex+f=0$. We need to find the relation between a, b, c, d, e and f.
We know that if three numbers p, q, and r are in G.P (Geometric Progression), then $q^2=pr$.
Since the numbers a, b, c are in G.P, we have $b^2=ac$.
$\Rightarrow b=\sqrt{ac}$ ---(1). We substitute this in the quadratic equation $a{x}^2+2bx+c=0$.
So, we have $a{x}^2+2\sqrt{ac}x+c=0$.
$\Rightarrow {\left(\sqrt{a}\right)}^2{x}^2+2\sqrt{ac}x+{\left(\sqrt{c}\right)}^2=0$.
$\Rightarrow {\left(\sqrt{a}\right)}^2{x}^2+\sqrt{ac}x+\sqrt{ac}x+{\left(\sqrt{c}\right)}^2=0$.
$\Rightarrow \sqrt{a}x(\sqrt{a}x+\sqrt{c})+\sqrt{c}(\sqrt{a}x+\sqrt{c})=0$.
$\Rightarrow (\sqrt{a}x+\sqrt{c})(\sqrt{a}x+\sqrt{c})=0$.
$\Rightarrow {(\sqrt{a}x+\sqrt{c})}^2=0$.
We have got both roots as equal.
$\Rightarrow (\sqrt{a}x+\sqrt{c})=0$.
$\Rightarrow \sqrt{a}x=-\sqrt{c}$.
$\Rightarrow x=-{\displaystyle \frac{\sqrt{c}}{\sqrt{a}}}$.
$$\Rightarrow x=-{\displaystyle \frac{\sqrt{c}\times \sqrt{a}}{\sqrt{a}\times \sqrt{a}}}$$.
$$\Rightarrow x=-{\displaystyle \frac{\sqrt{ac}}{\sqrt{a^2}}}$$.
From equation (1), we get.
$$\Rightarrow x=-{\displaystyle \frac{b}{a}}$$.
So, the root of the quadratic equation $a{x}^2+2bx+c=0$ is ${\displaystyle \frac{-b}{a}}$. Let us substitute this in the quadratic equation $d{x}^2+2ex+f=0$, as both the roots are equal.
So, we have $d{\left({\displaystyle \frac{-b}{a}}\right)}^2+2e\left({\displaystyle \frac{-b}{a}}\right)+f=0$.
$\Rightarrow \left({\displaystyle \frac{d{b}^2}{a^2}}\right)-\left({\displaystyle \frac{2eb}{a}}\right)+f=0$.
$\Rightarrow {\displaystyle \frac{d{b}^2-2aeb+f{a}^2}{a^2}}=0$.
Since denominator cannot be equal to zero, we get $d{b}^2-2aeb+f{a}^2=0$ ---(2).
From equation (1), we have $b^2=ac$. Let us substitute this in equation (2).
$\Rightarrow d\left(ac\right)-2aeb+f{a}^2=0$.
$\Rightarrow dac-2aeb+f{a}^2=0$.
$\Rightarrow dc-2eb+fa=0$.
Let us divide both sides with $ac$.
$\Rightarrow {\displaystyle \frac{dc-2eb+fa}{ac}}={\displaystyle \frac{0}{ac}}$.
$\Rightarrow {\displaystyle \frac{dc}{ac}}-{\displaystyle \frac{2eb}{ac}}+{\displaystyle \frac{fa}{ac}}=0$.
From equation (1), we get
$\Rightarrow {\displaystyle \frac{d}{a}}-{\displaystyle \frac{2eb}{b^2}}+{\displaystyle \frac{f}{c}}=0$.
$\Rightarrow {\displaystyle \frac{d}{a}}-{\displaystyle \frac{2e}{b}}+{\displaystyle \frac{f}{c}}=0$.
$\Rightarrow {\displaystyle \frac{d}{a}}+{\displaystyle \frac{f}{c}}=2\left({\displaystyle \frac{e}{b}}\right)$.
We know that if three numbers p, q, r are in A.P, then we have $p+r=2q$. Using this we can say that ${\displaystyle \frac{d}{a}}$, ${\displaystyle \frac{e}{b}}$, ${\displaystyle \frac{f}{c}}$ are in A.P.
∴ The correct option for the given problem is (c).

Note:
We can also solve the problem by assuming the common ratio of $a$, $b$, $c$ as $r$. We then makes substitutions $b=ar$, $c=a{r}^2$ in the quadratic equation $a{x}^2+2bx+c=0$ to find the roots of it. We then substitute those roots in quadratic equation $d{x}^2+2ex+f=0$ and make roots just as we did in the problem. We should not randomly take numbers for $a$, $b$, $c$ as this makes the calculation hectic.