Hard

Let $p$ and $q$ be the roots of the quadratic equation $ax^2+bx+c=0$, $a\ne0$. If $a$, $b$ and $c$ are in arithmetic progression and $\frac{1}{p}+\frac{1}{q}=4$, then the value of $(p-q)^2$ is

$\dfrac{48}{81}$

$\dfrac{61}{81}$

$\dfrac{68}{81}$

$\dfrac{52}{81}$

✅ Correct Option: 4

Using the formulas for the sum and product of roots of $ax^2+bx+c=0$: $p+q=-\dfrac{b}{a}$ and $pq=\dfrac{c}{a}$ The given condition becomes $\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{p+q}{pq}=\dfrac{-\frac{b}{a}}{\frac{c}{a}}=-\dfrac{b}{c}=4$ $\Rightarrow b=-4c$ Since $a$, $b$, $c$ are in arithmetic progression, the middle term is the average of the other two: $2b=a+c$ $\Rightarrow 2(-4c)=a+c$ $\Rightarrow a=-9c$ Now put these back: $p+q=-\dfrac{b}{a}=-\dfrac{-4c}{-9c}=-\dfrac{4}{9}$ $pq=\dfrac{c}{a}=\dfrac{c}{-9c}=-\dfrac{1}{9}$ Using the identity $(p-q)^2=(p+q)^2-4pq$: $(p-q)^2=\left(-\dfrac{4}{9}\right)^2-4\left(-\dfrac{1}{9}\right)$ $=\dfrac{16}{81}+\dfrac{4}{9}$ $=\dfrac{16}{81}+\dfrac{36}{81}$ $=\dfrac{52}{81}$

JIPMAT 2023

The graph of a polynomial $y = f(x)$ is shown in figure below, then the number of its zeros is:

JIPMAT 2024

The factors of polynomial $x^{3}-6 x^{2}+11 x-6$ are :

JIPMAT 2026

If one root of the quadratic equation $ax^2+bx+c=0$ is the reciprocal of other, then which of the following is correct?

JIPMAT 2024

If highest common factor of $x^{2}-p x-q$ and $5 x^{2}-3 p x-15 q$ is $(x-3)$ , then value of ( $\left.p, q\right)$ will be :

JIPMAT 2024

If $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$ , then value of

$\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}$ is :

JIPMAT 2024

If the roots of the equation $x^{2}-p x+54=0$ are in the ratio $2: 3$ , then value of $p$ is:

Let $p$ and $q$ be the roots of the quadratic equation $ax^2+bx+c=0$, $a\ne0$. If $a$, $b$ and $c$ are in arithmetic progression and $\frac{1}{p}+\frac{1}{q}=4$, then the value of $(p-q)^2$ is

Solution

Related questions:

Keyboard Shortcuts

JIPMAT Past Year Questions (Topic-Wise)