Q51 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics

If $\alpha$ and $\beta$ are the roots of the equation $2 x^2-4 x+3=0$, then $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=$

  • A
    $-1$
  • Correct answer
    B
    $-2$
  • C
    2
  • D
    1

Solution

$\because \alpha$ and $\beta$ are roots of $2 x^2-4 x+3=0$ $\Rightarrow \alpha+\beta=\frac{-(-4)}{2}=2$ and $\alpha \beta=\frac{3}{2}$ $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=4-2 \times \frac{3}{2}=1$ $\left(\alpha^2+\beta^2\right)^2=\alpha^4+\beta^4+2 \alpha^2 \beta^2$ $\Rightarrow \quad 1=\alpha^4+\beta^4+2 \times\left(\frac{3}{2}\right)^2$ $\Rightarrow \quad \alpha^4+\beta^4=1-\frac{9}{2} \Rightarrow \alpha^4+\beta^4=-\frac{7}{2}$ Now, $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=\frac{2 \times\left(-\frac{7}{2}\right)+3 \times 1}{2}$ $=\frac{-7+3}{2}=\frac{-4}{2}$ $\Rightarrow \frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=-2$

Hence, the correct answer is option: 'B'.

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