Question

Suppose $a,c$ are the roots of the equation $p{x}^2-3x+2=0$ and $b,d$ are the roots of the equation $q{x}^2-4x+2=0$. Find the value of $p$ and $q$ such that $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{d}$ are in A.P.

• A. $p=1,q=\frac{15}{8}$
• B. $p=0,q=\frac{3}{4}$
• C. $p=\frac{15}{8},q=\frac{5}{4}$
• D. $p=3,q=-1$

Answer 1

The correct option is A $p=1,q=\frac{15}{8}$

Given: $p{x}^2-3x+2=0$ with roots $a,c$ and $q{x}^2-4x+2=0$ with roots $b,d$ are quadratic equations and $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{d}$ are in A.P.

To find: $p$ and $q$

$\because a,c$ are the roots of $p{x}^2-3x+2=0$
$\Rightarrow a+c=\frac{3}{p}$ and $a.c=\frac{2}{p}$

Similarly $b,d$ are the roots of $q{x}^2-4x+2=0$
$\Rightarrow b+d=\frac{4}{q}$ and $b.d=\frac{2}{q}$

Since $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\Rightarrow \frac{2}{b}=\frac{a+c}{ac}\Rightarrow \frac{2}{b}=\frac{\frac{3}{p}}{\frac{2}{p}}=\frac{3}{2}$
$\Rightarrow b=\frac{4}{3}$

Also $\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in A.P.

$\Rightarrow \frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}\Rightarrow \frac{2}{c}=\frac{1}{b}+\frac{1}{d}$
$\Rightarrow \frac{2}{c}=\frac{b+d}{bd}\Rightarrow \frac{2}{c}=\frac{\frac{4}{q}}{\frac{2}{q}}=\frac{4}{2}=2$
$\Rightarrow c=\frac{2}{2}=1$

And $c$ is the roots of $p{x}^2-3x+2=0$

$\therefore p{c}^2-3c+2=0$
$\Rightarrow p(1{)}^2-3.1+2=0\Rightarrow p-3+2=0$
$\Rightarrow p-1=0\Rightarrow p=1$

Similarly, $b$ is the root of $q{x}^2-4x+2=0$
$\therefore q{b}^2-4b+2=0$
$\Rightarrow q{\left(\frac{4}{3}\right)}^2-4\left(\frac{4}{3}\right)+2=0$
$\Rightarrow \frac{16q}{9}-\frac{16}{3}+2=0$

$\Rightarrow \frac{16q-48+18}{9}=0$

$\Rightarrow 16q=30\Rightarrow q=\frac{15}{8}$

Hence, $p=1$ and $q=\frac{15}{8}$