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If $\alpha$ and  $\beta$ are the roots of the quadratic equation $x^{2}-10x+15= 0$, then find the quadratic equation whose roots are $\bigg(\alpha+\dfrac{\alpha }{\beta }\bigg)$ and $\bigg(\beta +\dfrac{\beta}{\alpha}\bigg)$

  1. $15x^{2}+71x+210= 0$
  2. $5x^{2}-22x+56= 0$
  3. $3x^{2}-44x+78= 0$
  4. Cannot be determined
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Given that$:$ $\alpha$ and  $\beta$ are the roots of the quadratic equation $x^{2} – 10x + 15 = 0$

Sum of roots $(\alpha + \beta) = \dfrac{10}{1} = 10$ and  Product of roots $(\alpha\cdot\beta) = \dfrac{15}{1} = 15$

Find the quadratic equation whose roots are $\left[\alpha+\dfrac{\alpha }{\beta }\right] \text{and} \left[\beta +\dfrac{\beta}{\alpha}\right]$

Sum of roots $ = \alpha+\dfrac{\alpha }{\beta } + \beta +\dfrac{\beta}{\alpha}$

$\implies \alpha + \beta + \dfrac{\alpha^{2} + \beta^{2}}{\alpha\beta}$

$\implies \alpha + \beta + \dfrac{(\alpha+ \beta)^{2}  – 2\alpha\beta}{\alpha\beta}$

$\implies 10 + \dfrac{100 – 30}{15} = \dfrac{220}{15} = \dfrac{44}{3}$

Product of roots $ = \left(\alpha+\dfrac{\alpha }{\beta }\right)\cdot \left(\beta +\dfrac{\beta}{\alpha}\right) = \alpha\beta + \beta + \alpha + 1 = 15 + 10 + 1 = 26$

Now, quadratic equation are $x^{2} – \left(\dfrac{44}{3}\right)x + 26 = 0$

$\implies 3x^{2} – 44x + 78 = 0$

So, the correct answer is $(C).$
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