JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Roots of a Quadratic Equation with Geometric Progression Coefficients

Q: Why does a cancel out completely from the equation?

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

Let α and β be the roots of the equation px^2 + qx - r = 0, where p ≠ 0. If p, q and r be the consecutive terms of a non-constant G.P and 1/α + 1/β = 3/4, then the value of (α - β)^2 is:

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A80/9B9C20/3D8

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 $(α - β)^{2} = \frac{80}{9}$ .

Option analysis

Why each option works or fails

A · 80/9

This is the correct value obtained from $(α - β)^{2} = (α + β)^{2} - 4 α β$ . Correct. From $\frac{α + β}{α β} = \frac{3}{4}$ , we find the common ratio $k = \frac{4}{3}$ , giving $(α - β)^{2} = k^{2} + 4 k^{2} = k^{2} + 4 k^{2}$ which evaluates to $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ .

B · 9

Taking $(α - β)^{2} = (α + β)^{2} - 4 α β$ 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 · 20/3

Writing $(α - β)^{2} = (α + β)^{2} - α β$ instead of subtracting $4 α β$ . The algebraic identity for difference squared in terms of sum and product is $(α - β)^{2} = (α + β)^{2} - 4 α β$ .

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 $p x^{2} + q x - r = 0$ ( $p \neq 0$ ) with roots $α, β$ . The coefficients $p, q, r$ form a non-constant G.P., and $\frac{1}{α} + \frac{1}{β} = \frac{3}{4}$ .

02Goal

Find the value of $(α - β)^{2}$ .

03Approach

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

04Execute

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

05Execute

Now compute $(α - β)^{2} = (α + β)^{2} - 4 α β = (- R)^{2} - 4 (- R^{2}) = R^{2} + 4 R^{2} = 5 R^{2} = 5 {(\frac{4}{3})}^{2} = 5 \times \frac{16}{9} = \frac{80}{9}$ .

✓Verify

Check that the discriminant of $x^{2} + \frac{4}{3} x - \frac{16}{9} = 0$ is $Δ = {(\frac{4}{3})}^{2} - 4 (1) (- \frac{16}{9}) = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which matches $(α - β)^{2} = \frac{Δ}{a_{l e a d}^{2}} = \frac{80}{9}$ .

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $(α - β)^{2} = \frac{80}{9}$ .

Why each option works or fails

A · correct: 80/9 - This is the correct value obtained from $(α - β)^{2} = (α + β)^{2} - 4 α β$ . Correct. From $\frac{α + β}{α β} = \frac{3}{4}$ , we find the common ratio $k = \frac{4}{3}$ , giving $(α - β)^{2} = k^{2} + 4 k^{2} = k^{2} + 4 k^{2}$ which evaluates to $\frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ .
B: 9 - Taking $(α - β)^{2} = (α + β)^{2} - 4 α β$ 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: 20/3 - Writing $(α - β)^{2} = (α + β)^{2} - α β$ instead of subtracting $4 α β$ . The algebraic identity for difference squared in terms of sum and product is $(α - β)^{2} = (α + β)^{2} - 4 α β$ .
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 $p x^{2} + q x - r = 0$ ( $p \neq 0$ ) with roots $α, β$ . The coefficients $p, q, r$ form a non-constant G.P., and $\frac{1}{α} + \frac{1}{β} = \frac{3}{4}$ .
Goal: Find the value of $(α - β)^{2}$ .
Approach: Represent the G.P. terms as $p = a$ , $q = a R$ , $r = a R^{2}$ with common ratio $R \neq 1$ . Reduce the quadratic to $x^{2} + R x - R^{2} = 0$ . Use Vieta's formulas to find $R$ from the sum of reciprocals, then compute $(α - β)^{2} = (α + β)^{2} - 4 α β$ .
Execute: Substitute $p = a, q = a R, r = a R^{2}$ into the quadratic equation: $a (x^{2} + R x - R^{2}) = 0 ⟹ x^{2} + R x - R^{2} = 0$ . By Vieta's relations, $α + β = - R$ and $α β = - R^{2}$ . Thus, $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- R}{- R^{2}} = \frac{1}{R} = \frac{3}{4} ⟹ R = \frac{4}{3}$ .
Execute: Now compute $(α - β)^{2} = (α + β)^{2} - 4 α β = (- R)^{2} - 4 (- R^{2}) = R^{2} + 4 R^{2} = 5 R^{2} = 5 {(\frac{4}{3})}^{2} = 5 \times \frac{16}{9} = \frac{80}{9}$ .
Verify: Check that the discriminant of $x^{2} + \frac{4}{3} x - \frac{16}{9} = 0$ is $Δ = {(\frac{4}{3})}^{2} - 4 (1) (- \frac{16}{9}) = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$ , which matches $(α - β)^{2} = \frac{Δ}{a_{l e a d}^{2}} = \frac{80}{9}$ .

Roots of a Quadratic Equation with Geometric Progression Coefficients

What feels right?

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

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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