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(B) Given the quadratic equation $2x^2 - 35x + 2 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = \frac{35}{2}$ and

Vedclass · Accepted Answer

$64$

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(B) Given the quadratic equation $2x^2 - 35x + 2 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = \frac{35}{2}$ and $\alpha\beta = \frac{2}{2} = 1$.Since $\alpha$ is a root,$2\alpha^2 - 35\alpha + 2 = 0$,which implies $2\alpha^2 - 35\alpha = -2$.Dividing by $\alpha$,we get $2\alpha - 35 = -\frac{2}{\alpha}$.Similarly,for root $\beta$,$2\beta - 35 = -\frac{2}{\beta}$.Now,we need to evaluate $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$.Substituting the expressions derived above:$(2\alpha - 35)^3 \cdot (2\beta - 35)^3 = \left(-\frac{2}{\alpha}\right)^3 \cdot \left(-\frac{2}{\beta}\right)^3$.$= \left(-\frac{8}{\alpha^3}\right) \cdot \left(-\frac{8}{\beta^3}\right) = \frac{64}{(\alpha\beta)^3}$.Since $\alpha\beta = 1$,the expression becomes $\frac{64}{1^3} = 64$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 35x + 2 = 0$,then the value of $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$ is equal to ......

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