Question

If p, q, r in harmonic progression and p and r be different having the same sign then the roots of the equation $p{{x}^{2}}+qx+r=0$ are
(a) Real and equal
(b) Real and distinct
(c) Irrational
(d) Imaginary

Answer 1

Hint: First, before we proceed further, we must know the formula of the harmonic progression(HP) with terms p, q, and r as $q=\dfrac{2pr}{p+r}$. Then, by using the quadratic formula to get the roots of the equation given as $p{{x}^{2}}+qx+r=0$ is given by the formula as $\sqrt{{{q}^{2}}-4pr}$. Then, substituting the value of q in the above expression, we get the condition for the roots of the given equation.

Complete step-by-step solution:
In this question, we are supposed to find the roots of the equation $p{{x}^{2}}+qx+r=0$ with the given condition p, q, r in harmonic progression and p and r be different having the same sign.
Now, before we proceed further, we must know the formula of the harmonic progression(HP) with terms p, q and r as:
$q=\dfrac{2pr}{p+r}$
Then, by using the quadratic formula to get the roots of the equation a given as $p{{x}^{2}}+qx+r=0$ is given by the formula as:
$\sqrt{{{q}^{2}}-4pr}$
Now, by substituting the value of q in the above expression as:
$\sqrt{{{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr}$
Now, we have the condition in the question that p and r is having the same value, so let be both positive, then we get:
$\begin{align}
  & \sqrt{{{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr}<0 \\
 & \Rightarrow {{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr<0 \\
\end{align}$
Now, by using the condition again in the question that p and r is having the same value, so let be both negative, then we get:
$\begin{align}
  & \sqrt{{{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr}<0 \\
 & \Rightarrow {{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr<0 \\
\end{align}$
So, both the times, we get the value of the discriminate which is ${{\left( \dfrac{2pr}{p+r} \right)}^{2}}-4pr$ less than zero which proves, the equation has imaginary roots.
So, the above condition clearly states that if it is less than zero, then the roots of the equation $p{{x}^{2}}+qx+r=0$ is imaginary.
Hence, option (d) is correct.

Note: Now, to solve these type of the questions we need to know some of the basic conditions for the roots of the equation $p{{x}^{2}}+qx+r=0$ is given as:
Roots of the equation is imaginary if $\sqrt{{{q}^{2}}-4pr}<0$.
Roots of the equation is real and equal if $\sqrt{{{q}^{2}}-4pr}=0$.
Roots of the equation is real and distinct if $\sqrt{{{q}^{2}}-4pr}>0$.