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(B) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 - 35x + 2 = 0$.From the properties of roots,the product of roots $\a

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(B) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 - 35x + 2 = 0$.From the properties of roots,the product of roots $\alpha \beta = \frac{c}{a} = \frac{2}{2} = 1$.Since $\alpha$ is a root,it satisfies the equation: $2\alpha^2 - 35\alpha + 2 = 0$,which implies $2\alpha^2 - 35\alpha = -2$.Dividing by $\alpha$ (assuming $\alpha 
eq 0$),we get $2\alpha - 35 = -\frac{2}{\alpha}$.Similarly,for root $\beta$,we have $2\beta - 35 = -\frac{2}{\beta}$.Now,substitute these into the expression $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$:$(2\alpha - 35)^3 \cdot (2\beta - 35)^3 = \left(-\frac{2}{\alpha}ight)^3 \cdot \left(-\frac{2}{\beta}ight)^3$.$= \left(-\frac{8}{\alpha^3}ight) \cdot \left(-\frac{8}{\beta^3}ight) = \frac{64}{(\alpha \beta)^3}$.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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