Question

A quadratic function $y=p{x}^2+qx+r$, where $p,q$ and $r$ are distinct real numbers satisfies the below conditions :
$\left(1\right)$ The graph of $y$ passes through the point $(1,6)$.
$\left(2\right)\frac{1}{p},\frac{1}{q},\frac{1}{r}$ form an arithmetic progression.
$\left(3\right)p,r,q$ are in geometric progression.

Which of the following options is/are correct?

• A. Sum of the roots of $y=0$ is $-4$.
• B. Product of the roots of $y=0$ is $-2.$
• C. There exist $2$ real and distinct roots of $y=0$
• D. There exist $2$ real and equal roots of $y=0$.

Answer 1

Answer: (A), (B), (C)

There exist $2$ real and distinct roots of $y=0$

Given : $y=p{x}^2+qx+r$
This passes through $(1,6)$, so
$p+q+r=6\cdots \left(1\right)$
$p,r,q$ are in geometric progression.
Let the common ratio be $c$, then
$r=pc,q=p{c}^2\cdots \left(2\right)$
$\frac{1}{p},\frac{1}{q},\frac{1}{r}$ form an arithmetic progression, then
$\frac{1}{p}+\frac{1}{r}=\frac{2}{q}$
Using equation $\left(2\right)$, we get
$1+\frac{1}{c}=\frac{2}{c^2}\Rightarrow c^2+c-2=0\Rightarrow (c+2)(c-1)=0\Rightarrow c=-2,1$

When $c=1$, then $p=q=r$, so it is rejected.
$\therefore c=-2\Rightarrow r=-2p,q=4p$
From equation $\left(1\right)$, we get
$p+4p-2p=6\Rightarrow p=2\Rightarrow q=8,r=-4\therefore y=2x^2+8x-4\Rightarrow y=2(x^2+4x-2)D=16+8=24>0$
$2$ real and distinct roots.
Sum of roots $=-4$
Product of roots $=-2$