If the roots of the equation 8{x^3} - 14{x^2} | vedclass

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Question

(a) Let the roots be $\frac{\alpha }{\beta },\alpha ,\alpha \beta ,\beta 
e 0$.

Then the product of roots is ${\alpha ^3} = - \frac{{ - 1}}{8} = \

Vedclass · Accepted Answer

$1,\frac{1}{2},\frac{1}{4}$

Vedclass · Answer

(a) Let the roots be $\frac{\alpha }{\beta },\alpha ,\alpha \beta ,\beta 
e 0$.

Then the product of roots is ${\alpha ^3} = - \frac{{ - 1}}{8} = \frac{1}{8} \Rightarrow \,\,\alpha = \frac{1}{2}$ and hence $\beta = \frac{1}{2}$,

so roots are $1,\frac{1}{2},\frac{1}{4}$

Trick : By inspection, we get the numbers $1,\frac{1}{2},\frac{1}{4}$ satisfying the given equation.

If the roots of the equation $8{x^3} - 14{x^2} + 7x - 1 = 0$ are in $G.P.$, then the roots are

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