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) $\dfrac{d}{a}$, $\dfrac{e}{b}$, $\dfrac{f}{c}$ are in G.P.
(c) $\dfrac{d}{a}$, $\dfrac{e}{b}$, $\dfrac{f}{c}$ 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 $a{{x}^{2}}+2\sqrt{ac}x+c=0$.
$\Rightarrow {{\left( \sqrt{a} \right)}^{2}}{{x}^{2}}+2\sqrt{ac}x+{{\left( \sqrt{c} \right)}^{2}}=0$.
$\Rightarrow {{\left( \sqrt{a} \right)}^{2}}{{x}^{2}}+\sqrt{ac}x+\sqrt{ac}x+{{\left( \sqrt{c} \right)}^{2}}=0$.
$\Rightarrow \sqrt{a}x\left( \sqrt{a}x+\sqrt{c} \right)+\sqrt{c}\left( \sqrt{a}x+\sqrt{c} \right)=0$.
$\Rightarrow \left( \sqrt{a}x+\sqrt{c} \right)\left( \sqrt{a}x+\sqrt{c} \right)=0$.
$\Rightarrow {{\left( \sqrt{a}x+\sqrt{c} \right)}^{2}}=0$.
We have got both roots as equal.
$\Rightarrow \left( \sqrt{a}x+\sqrt{c} \right)=0$.
$\Rightarrow \sqrt{a}x=-\sqrt{c}$.
$\Rightarrow x=-\dfrac{\sqrt{c}}{\sqrt{a}}$.
\[\Rightarrow x=-\dfrac{\sqrt{c}\times \sqrt{a}}{\sqrt{a}\times \sqrt{a}}\].
\[\Rightarrow x=-\dfrac{\sqrt{ac}}{\sqrt{{{a}^{2}}}}\].
From equation (1), we get.
\[\Rightarrow x=-\dfrac{b}{a}\].
So, the root of the quadratic equation $a{{x}^{2}}+2bx+c=0$ is $\dfrac{-b}{a}$. Let us substitute this in the quadratic equation $d{{x}^{2}}+2ex+f=0$, as both the roots are equal.
So, we have $d{{\left( \dfrac{-b}{a} \right)}^{2}}+2e\left( \dfrac{-b}{a} \right)+f=0$.
$\Rightarrow \left( \dfrac{d{{b}^{2}}}{{{a}^{2}}} \right)-\left( \dfrac{2eb}{a} \right)+f=0$.
$\Rightarrow \dfrac{d{{b}^{2}}-2aeb+f{{a}^{2}}}{{{a}^{2}}}=0$.
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 $ac$.
$\Rightarrow \dfrac{dc-2eb+fa}{ac}=\dfrac{0}{ac}$.
$\Rightarrow \dfrac{dc}{ac}-\dfrac{2eb}{ac}+\dfrac{fa}{ac}=0$.
From equation (1), we get
$\Rightarrow \dfrac{d}{a}-\dfrac{2eb}{{{b}^{2}}}+\dfrac{f}{c}=0$.
$\Rightarrow \dfrac{d}{a}-\dfrac{2e}{b}+\dfrac{f}{c}=0$.
$\Rightarrow \dfrac{d}{a}+\dfrac{f}{c}=2\left( \dfrac{e}{b} \right)$.
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 $\dfrac{d}{a}$, $\dfrac{e}{b}$, $\dfrac{f}{c}$ 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.