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If $\alpha$ and $\beta$ are the roots of the equation
$2 x^2-4 x+3=0$, then $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=$
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$2 x^2-4 x+3=0$, then $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=$
Solution:
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Verified Answer
The correct answer is:
$-2$
$\because \alpha$ 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, $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=\frac{2 \times\left(-\frac{7}{2}\right)+3 \times 1}{2}$
$=\frac{-7+3}{2}=\frac{-4}{2}$
$\Rightarrow \frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=-2$
$\Rightarrow \alpha+\beta=\frac{-(-4)}{2}=2$ 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, $\frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=\frac{2 \times\left(-\frac{7}{2}\right)+3 \times 1}{2}$
$=\frac{-7+3}{2}=\frac{-4}{2}$
$\Rightarrow \frac{2\left(\alpha^4+\beta^4\right)+3\left(\alpha^2+\beta^2\right)}{\alpha+\beta}=-2$
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