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(D) For the quadratic equation $x^{2}+3ax+2a^{2}=0$,the sum of the roots $\alpha+\beta = -\frac{b}{a_{coeff}} = -3a$ and the product of the roots $\al

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$\pm 1$

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(D) For the quadratic equation $x^{2}+3ax+2a^{2}=0$,the sum of the roots $\alpha+\beta = -\frac{b}{a_{coeff}} = -3a$ and the product of the roots $\alpha\beta = \frac{c}{a_{coeff}} = 2a^{2}$.We know that $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.Given $\alpha^{2}+\beta^{2} = 5$,we substitute the values:$5 = (-3a)^{2} - 2(2a^{2})$$5 = 9a^{2} - 4a^{2}$$5 = 5a^{2}$$a^{2} = 1$$a = \pm 1$.

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+3ax+2a^{2}=0$ and if $\alpha^{2}+\beta^{2}=5$,find the value of $a$.

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