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

• A. $3x^2+19x+3=0$
• B. $3x^2+19x-3=0$
• C. $3x^2-19x+3=0$
• D. $3x^2-19x-3=0$

Hint:

For roots involving ratios, first find the sum and product of the original roots using Vieta's formulas.

Answer 1

The Correct Option is C

Step 1: Find the equation $\alpha,\beta$ satisfy.
We are told $\alpha^2=5\alpha-3$ and $\beta^2=5\beta-3$. So both $\alpha$ and $\beta$ are roots of $x^2-5x+3=0.$
Step 2: Get sum and product.
From $x^2-5x+3=0$: sum $\alpha+\beta=5$ and product $\alpha\beta=3.$
Step 3: Name the new roots.
The new equation has roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$. Call them $r_1,r_2$. We need their sum and product.
Step 4: Product of new roots.
$r_1r_2=\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha}=1.$
Step 5: Sum of new roots.
$r_1+r_2=\dfrac{\alpha^2+\beta^2}{\alpha\beta}.$ Now $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=25-6=19.$ So $r_1+r_2=\dfrac{19}{3}.$
Step 6: Build the equation.
Using $x^2-(\text{sum})x+(\text{product})=0$: $x^2-\dfrac{19}{3}x+1=0.$ Multiply by $3$: \[ 3x^2-19x+3=0. \] \[ \boxed{3x^2-19x+3=0} \]