Question:
If $\alpha \neq \beta$ and $\alpha^2=5\alpha-3,\ \beta^2=5\beta-3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is
If $\alpha \neq \beta$ and $\alpha^2=5\alpha-3,\ \beta^2=5\beta-3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is
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For roots involving ratios, first find the sum and product of the original roots using Vieta's formulas.
Updated On: Jun 3, 2026
- $3x^2+19x+3=0$
- $3x^2+19x-3=0$
- $3x^2-19x+3=0$
- $3x^2-19x-3=0$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
First determine $\alpha+\beta$ and $\alpha\beta$ from the quadratic satisfied by $\alpha$ and $\beta$.
Step 2: Meaning
Since \[ x^2-5x+3=0 \] has roots $\alpha,\beta$, \[ \alpha+\beta=5,\qquad \alpha\beta=3. \]
Step 3: Analysis
Let \[ r_1=\frac{\alpha}{\beta},\qquad r_2=\frac{\beta}{\alpha}. \] Then \[ r_1r_2=1. \] Also, \[ r_1+r_2 = \frac{\alpha^2+\beta^2}{\alpha\beta}. \] Now \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta =25-6=19. \] Hence \[ r_1+r_2=\frac{19}{3}. \]
Step 4: Conclusion
Required equation: \[ x^2-\frac{19}{3}x+1=0. \] Multiplying by $3$, \[ 3x^2-19x+3=0. \]
Final Answer: (C)
First determine $\alpha+\beta$ and $\alpha\beta$ from the quadratic satisfied by $\alpha$ and $\beta$.
Step 2: Meaning
Since \[ x^2-5x+3=0 \] has roots $\alpha,\beta$, \[ \alpha+\beta=5,\qquad \alpha\beta=3. \]
Step 3: Analysis
Let \[ r_1=\frac{\alpha}{\beta},\qquad r_2=\frac{\beta}{\alpha}. \] Then \[ r_1r_2=1. \] Also, \[ r_1+r_2 = \frac{\alpha^2+\beta^2}{\alpha\beta}. \] Now \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta =25-6=19. \] Hence \[ r_1+r_2=\frac{19}{3}. \]
Step 4: Conclusion
Required equation: \[ x^2-\frac{19}{3}x+1=0. \] Multiplying by $3$, \[ 3x^2-19x+3=0. \]
Final Answer: (C)
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