Question:
If the roots of the equation
\[
x^3-px^2+qx-s=0
\]
are in geometric progression, then
If the roots of the equation
\[
x^3-px^2+qx-s=0
\]
are in geometric progression, then
Show Hint
For roots in G.P., use $\frac{a}{r},a,ar$ to simplify Vieta relations quickly.
Updated On: Jun 3, 2026
- $q^2=ps^2$
- $q^2=p^2s$
- $q^3=ps^3$
- $q^3=p^3s$
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Represent the roots in geometric progression.
Step 2: Meaning
Let the roots be \[ \frac{a}{r},\ a,\ ar. \]
Step 3: Analysis
Using Vieta's formulas, \[ p=\frac{a}{r}+a+ar =a\left(r+1+\frac1r\right), \] \[ q=\frac{a^2}{r}+a^2+a^2r =a^2\left(r+1+\frac1r\right), \] and \[ s=\frac{a}{r}\cdot a\cdot ar=a^3. \] Therefore, \[ q=a\,p. \] Cubing both sides, \[ q^3=a^3p^3. \] Since $a^3=s$, \[ q^3=p^3s. \]
Step 4: Conclusion
Hence the required relation is \[ q^3=p^3s. \]
Final Answer: (D)
Represent the roots in geometric progression.
Step 2: Meaning
Let the roots be \[ \frac{a}{r},\ a,\ ar. \]
Step 3: Analysis
Using Vieta's formulas, \[ p=\frac{a}{r}+a+ar =a\left(r+1+\frac1r\right), \] \[ q=\frac{a^2}{r}+a^2+a^2r =a^2\left(r+1+\frac1r\right), \] and \[ s=\frac{a}{r}\cdot a\cdot ar=a^3. \] Therefore, \[ q=a\,p. \] Cubing both sides, \[ q^3=a^3p^3. \] Since $a^3=s$, \[ q^3=p^3s. \]
Step 4: Conclusion
Hence the required relation is \[ q^3=p^3s. \]
Final Answer: (D)
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