Why does cancel out completely from the equation?

Because , , so dividing the entire quadratic equation by gives an equivalent monic quadratic equation depending only on .

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Let

\alpha

and

\beta

be the roots of the equation

\mathrm{p}x^2 + \mathrm{q}x - \mathrm{r} = 0

, where

\mathrm{p} \neq 0

. If

\mathrm{p}, \mathrm{q}

and

\mathrm{r}

be the consecutive terms of a non-constant G.P and

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}

, then the value of

(\alpha - \beta)^2

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{80}{9}

9

\frac{20}{3}

8

Step-by-step solutionView

Correct answer

Using Vieta's formulas along with the geometric progression relation gives the common ratio $k = 4/3$ , leading to $(\alpha - \beta)^2 = \frac{80}{9}$ .

Option analysis

Why each option works or fails

A ·

\frac{80}{9}

None. This is the correct value obtained from $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ . Correct. From $\frac{\alpha+\beta}{\alpha\beta} = \frac{3}{4}$ , we find the common ratio $k = \frac{4}{3}$ , giving $(\alpha - \beta)^2 = k^2 + 4k^2 = k^2 + 4k^2$ which evaluates to $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ .

B ·

9

Taking $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ but forgetting the product term's coefficient or miscalculating the common ratio as $k=3$ . Check the fraction arithmetic when evaluating $\frac{1}{k} = \frac{3}{4}$ , which yields $k = \frac{4}{3}$ , not an integer.

C ·

\frac{20}{3}

Writing $(\alpha - \beta)^2 = (\alpha + \beta)^2 - \alpha\beta$ instead of subtracting $4\alpha\beta$ . The algebraic identity for difference squared in terms of sum and product is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ .

D ·

8

Dropping the denominator $9$ from the squared ratio during final simplification. Keep all fractional terms with a common denominator of $9$ : $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which does not simplify to an integer.

Reviewed route

Solution

StepWorking

01given

The quadratic equation is $px^2 + qx - r = 0$ ( $p \neq 0$ ) with roots $\alpha, \beta$ . The coefficients $p, q, r$ form a non-constant G.P., and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$ .

02goal

Find the value of $(\alpha - \beta)^2$ .

03approach

Represent the G.P. terms as $p = a$ , $q = aR$ , $r = aR^2$ with common ratio $R \neq 1$ . Reduce the quadratic to $x^2 + Rx - R^2 = 0$ . Use Vieta's formulas to find $R$ from the sum of reciprocals, then compute $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ .

04execute

Substitute $p = a, q = aR, r = aR^2$ into the quadratic equation: $a(x^2 + Rx - R^2) = 0 \implies x^2 + Rx - R^2 = 0$ . By Vieta's relations, $\alpha + \beta = -R$ and $\alpha\beta = -R^2$ . Thus, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-R}{-R^2} = \frac{1}{R} = \frac{3}{4} \implies R = \frac{4}{3}$ .

05execute

Now compute $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (-R)^2 - 4(-R^2) = R^2 + 4R^2 = 5R^2 = 5\left(\frac{4}{3}\right)^2 = 5 \times \frac{16}{9} = \frac{80}{9}$ .

✓verify

Check that the discriminant of $x^2 + \frac{4}{3}x - \frac{16}{9} = 0$ is $\Delta = \left(\frac{4}{3}\right)^2 - 4(1)\left(-\frac{16}{9}\right) = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which matches $(\alpha - \beta)^2 = \frac{\Delta}{a_{lead}^2} = \frac{80}{9}$ .

Hints that build this answer step by step

For the quadratic equation $\mathrm{p}x^2 + \mathrm{q}x - \mathrm{r} = 0$ , how are the sum $\alpha + \beta$ and product $\alpha\beta$ expressed in terms of the coefficients?
$\alpha + \beta = -\frac{\mathrm{q}}{\mathrm{p}}$ and $\alpha\beta = -\frac{\mathrm{r}}{\mathrm{p}}$
Let $\mathrm{p}, \mathrm{q}, \mathrm{r}$ be in G.P. with common ratio $k$ such that $\mathrm{q} = \mathrm{p}k$ and $\mathrm{r} = \mathrm{p}k^2$ . Using $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$ , what is the value of $k$ ?
$k = \frac{4}{3}$
What is the value of $(\alpha - \beta)^2$ using $k = \frac{4}{3}$ ?
$\frac{80}{9}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does

a

cancel out completely from the equation?

Because $p \neq 0$ , $a \neq 0$ , so dividing the entire quadratic equation by $a$ gives an equivalent monic quadratic equation depending only on $R$ .

Answer

Using Vieta's formulas along with the geometric progression relation gives the common ratio $k = 4/3$ , leading to $(\alpha - \beta)^2 = \frac{80}{9}$ .

Why each option works or fails

A · correct: $\frac{80}{9}$ - None. This is the correct value obtained from $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ . Correct. From $\frac{\alpha+\beta}{\alpha\beta} = \frac{3}{4}$ , we find the common ratio $k = \frac{4}{3}$ , giving $(\alpha - \beta)^2 = k^2 + 4k^2 = k^2 + 4k^2$ which evaluates to $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ .
B: $9$ - Taking $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ but forgetting the product term's coefficient or miscalculating the common ratio as $k=3$ . Check the fraction arithmetic when evaluating $\frac{1}{k} = \frac{3}{4}$ , which yields $k = \frac{4}{3}$ , not an integer.
C: $\frac{20}{3}$ - Writing $(\alpha - \beta)^2 = (\alpha + \beta)^2 - \alpha\beta$ instead of subtracting $4\alpha\beta$ . The algebraic identity for difference squared in terms of sum and product is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ .
D: $8$ - Dropping the denominator $9$ from the squared ratio during final simplification. Keep all fractional terms with a common denominator of $9$ : $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which does not simplify to an integer.

Step-by-step solution

given: The quadratic equation is $px^2 + qx - r = 0$ ( $p \neq 0$ ) with roots $\alpha, \beta$ . The coefficients $p, q, r$ form a non-constant G.P., and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$ .
goal: Find the value of $(\alpha - \beta)^2$ .
approach: Represent the G.P. terms as $p = a$ , $q = aR$ , $r = aR^2$ with common ratio $R \neq 1$ . Reduce the quadratic to $x^2 + Rx - R^2 = 0$ . Use Vieta's formulas to find $R$ from the sum of reciprocals, then compute $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ .
execute: Substitute $p = a, q = aR, r = aR^2$ into the quadratic equation: $a(x^2 + Rx - R^2) = 0 \implies x^2 + Rx - R^2 = 0$ . By Vieta's relations, $\alpha + \beta = -R$ and $\alpha\beta = -R^2$ . Thus, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-R}{-R^2} = \frac{1}{R} = \frac{3}{4} \implies R = \frac{4}{3}$ .
execute: Now compute $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (-R)^2 - 4(-R^2) = R^2 + 4R^2 = 5R^2 = 5\left(\frac{4}{3}\right)^2 = 5 \times \frac{16}{9} = \frac{80}{9}$ .
verify: Check that the discriminant of $x^2 + \frac{4}{3}x - \frac{16}{9} = 0$ is $\Delta = \left(\frac{4}{3}\right)^2 - 4(1)\left(-\frac{16}{9}\right) = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which matches $(\alpha - \beta)^2 = \frac{\Delta}{a_{lead}^2} = \frac{80}{9}$ .

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

Using Vieta's formulas along with the geometric progression relation gives the common ratio $k = 4/3$ , leading to $(\alpha - \beta)^2 = \frac{80}{9}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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