Question

If $a,b$ and $c$ are positive numbers in a GP, then the roots of the quadratic equation $({\log }_{e}a)x^2-2({\log }_{e}b)x+({\log }_{e}c)=0$ are

• A. $–1$ and $\frac{[{\log }_{e}c]}{[{\log }_{e}a]}$
• B. $1$ and $\frac{[{\log }_{e}c]}{[{\log }_{e}a]}$
• C. $1$ and ${\log }_{a}c$
• D. $–1$ and ${\log }_{c}a$

Answer 1

The correct option is C

$1$ and ${\log }_{a}c$

Explanation for the correct option:

Step 1. Find the roots of given quadratic equation:

$\because a,b$ and $c$ are in G.P

$\therefore b^2=ac$

Given equation is $({\log }_{e}a)x^2-2({\log }_{e}b)x+({\log }_{e}c)=0$

Step 2. Put $x=1$ in the given equation, we get

$({\log }_{e}a)-2({\log }_{e}b)+({\log }_{e}c)=0$

$\Rightarrow$$2({\log }_{e}b)=({\log }_{e}a)+({\log }_{e}c)$

$\Rightarrow$ ${\log }_e{b}^2={\log }_{e}ac$ $\left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{A}\mathit{B}\right]$

$\Rightarrow$ $b^2=ac$, which is true

Hence, one of the root of this equation is $1$

Step 3. Let another root be $\alpha$:

$\therefore$Sum of the roots, $1+\alpha =\frac{2{\log }_{e}b}{{\log }_{e}a}$

$=\frac{{\log }_e{b}^2}{{\log }_{e}a}$

$\begin{array}{rcl}\alpha & =& \frac{{\log }_{e}ac}{{\log }_{e}a}-1\\ & =& \frac{({\log }_{e}a+{\log }_{e}c)}{{\log }_{e}a}-1\\ & =& \frac{{\log }_{e}c}{{\log }_{e}a}\\ & =& {\log }_{a}c\end{array}$

Thus, the roots of given quadratic equation are $\left(1,lo{g}_{a}c\right)$

Hence, Option ‘C’ is Correct.