Sequences and Series 3 Question 2
2. If three distinct numbers $a, b$ and $c$ are in GP and the equations $a x^{2}+2 b x+c=0$ and $d x^{2}+2 e x+f=0$ have a common root, then which one of the following statements is correct?
(2019 Main, 8 April II)
(a) $d, e$ and $f$ are in GP
(b) $\frac{d}{a}, \frac{e}{b}$ and $\frac{f}{c}$ are in AP
(c) $d, e$ and $f$ are in AP
(d) $\frac{d}{a}, \frac{e}{b}$ and $\frac{f}{c}$ are in GP
Show Answer
Answer:
Correct Answer: 2. (b)
Solution:
- (b) Given, three distinct numbers $a, b$ and $c$ are in GP. $\therefore \quad b^{2}=a c$ and the given quadratic equations
$$ \begin{aligned} & a x^{2}+2 b x+c=0 \\ & d x^{2}+2 e x+f=0 \end{aligned} $$
For quadratic Eq. (ii), the discriminant $D=(2 b)^{2}-4 a c$
$$ =4\left(b^{2}-a c\right)=0 $$
$\Rightarrow$ Quadratic Eq. (ii) have equal roots, and it is equal to $x=-\frac{b}{a}$, and it is given that quadratic Eqs. (ii) and (iii) have a common root, so
$$ \begin{array}{rcr} & d-\frac{b}{a}^{2} \quad+2 e-\frac{b}{a}+f=0 & \\ \Rightarrow & d b^{2}-2 e b a+a^{2} f=0 & {\left[\because b^{2}=a c\right]} \\ \Rightarrow & d(a c)-2 e a b+a^{2} f=0 & {[\because a \neq 0]} \\ \Rightarrow & d c-2 e b+a f=0 & \\ \Rightarrow & 2 e b=d c+a f & \\ \Rightarrow & 2 \frac{e}{b}=\frac{d c}{b^{2}}+\frac{a f}{b^{2}} \\ & 2 \frac{e}{b}=\frac{d}{a}+\frac{f}{c} & {\left[\because b^{2}=a c\right]} \end{array} $$
So, $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in AP.
Alternate Solution
Given, three distinct numbers $a, b$ and $c$ are in GP. Let $a=a, b=a r, c=a r^{2}$ are in GP, which satisfies $a x^{2}+2 b x+c=0$
$$ \begin{array}{ll} \therefore & a x^{2}+2(a r) x+a r^{2}=0 \\ \Rightarrow & x^{2}+2 r x+r^{2}=0 \\ \Rightarrow & (x+r)^{2}=0 \Rightarrow x=-r . \end{array} $$
According to the question, $a x^{2}+2 b x+c=0$ and $d x^{2}+2 e x+f=0$ have a common root.
So, $x=-r$ satisfies $d x^{2}+2 e x+f=0$
$\therefore \quad d(-r)^{2}+2 e(-r)+f=0$
$\Rightarrow \quad d r^{2}-2 e r+f=0$
$\Rightarrow \quad d \frac{c}{a}-2 e \frac{c}{b}+f=0$
$\Rightarrow \quad \frac{d}{a}-\frac{2 e}{b}+\frac{f}{c}=0$
$\Rightarrow \quad \frac{d}{a}+\frac{f}{c}=\frac{2 e}{b}$
$[\because c \neq 0]$
