The value of c for which |{alpha ^2} - {beta | vedclass

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(C) Given the quadratic equation $2{x^2} + 7x + c = 0$. From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{7}{2}$ and $

Vedclass · Accepted Answer

$6$

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(C) Given the quadratic equation $2{x^2} + 7x + c = 0$. From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{7}{2}$ and $\alpha \beta = \frac{c}{2}$. We are given $|{\alpha ^2} - {\beta ^2}| = \frac{7}{4}$,which implies $|(\alpha + \beta)(\alpha - \beta)| = \frac{7}{4}$. Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we have $(\alpha - \beta)^2 = (-\frac{7}{2})^2 - 4(\frac{c}{2}) = \frac{49}{4} - 2c$. Thus,$|\alpha - \beta| = \sqrt{\frac{49 - 8c}{4}} = \frac{\sqrt{49 - 8c}}{2}$. Substituting these into the equation: $|-\frac{7}{2}| 	imes \frac{\sqrt{49 - 8c}}{2} = \frac{7}{4}$. $\frac{7}{2} 	imes \frac{\sqrt{49 - 8c}}{2} = \frac{7}{4}$. $\frac{7\sqrt{49 - 8c}}{4} = \frac{7}{4}$. $\sqrt{49 - 8c} = 1$. Squaring both sides,$49 - 8c = 1$. $8c = 48$,so $c = 6$.

The value of $c$ for which $|{\alpha ^2} - {\beta ^2}| = \frac{7}{4}$,where $\alpha$ and $\beta$ are the roots of $2{x^2} + 7x + c = 0$,is

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