Question

If $\alpha ,\beta$ are the roots of the equation $8x^2-3x+27=0$, then the value of $\left[{\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}\right]$ is:

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{1}{4}$

Step 1: Find the sum and product of the roots of the quadratic equation.

It is given that $\alpha ,\beta$ are the roots of the equation $8x^2-3x+27=0$.

Since the given equation is in the standard form of a quadratic equation $a{x}^2+bx+c=0$.

Therefore, the sum of the roots is given by $\alpha +\beta =-\frac{b}{a}$, and the product of the roots is $\alpha ·\beta =\frac{c}{a}$.

For the equation $8x^2-3x+27=0$, the sum of the roots is given as follows:

$\begin{array}{rcl}\alpha +\beta & =& \frac{3}{8}...\left(1\right)\end{array}$

And, the product of the roots is:

$\begin{array}{rcl}\alpha \beta & =& \frac{27}{8}...\left(2\right)\end{array}$

Step 2: Simplify the given expression.

Now, simplify the expression $\left[{\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}\right]$ as follows:

$\begin{array}{rcl}\left[{\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}\right]& =& \frac{{\left({\alpha }^2\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(\beta \right)}^{{\displaystyle \frac{1}{3}}}}+\frac{{\left({\beta }^2\right)}^{\frac{1}{3}}}{{\left(\alpha \right)}^{\frac{1}{3}}}\\ & =& \frac{{\left(\alpha \right)}^{\frac{2}{3}}{\left(\alpha \right)}^{\frac{1}{3}}+{\left(\beta \right)}^{\frac{2}{3}}{\left(\beta \right)}^{\frac{1}{3}}}{{\left(\alpha \beta \right)}^{\frac{1}{3}}}\\ & =& \frac{\alpha +\beta }{{\left(\alpha \beta \right)}^{\frac{1}{3}}}\end{array}$

Substitute the values of $\alpha +\beta$ and $\alpha \beta$ from equation (1) and equation (2) into the above equation and simplify.

$\begin{array}{rcl}\frac{\alpha +\beta }{{\left(\alpha \beta \right)}^{\frac{1}{3}}}& =& \frac{{\displaystyle \frac{3}{8}}}{{\left({\displaystyle \frac{27}{8}}\right)}^{{\displaystyle \frac{1}{3}}}}\\ & =& \frac{{\displaystyle \frac{3}{8}}}{{\displaystyle \frac{3}{2}}}\\ & =& \frac{1}{4}\end{array}$

Hence, the value of $\left[{\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}\right]$ is $\frac{1}{4}$.

Hence, option B is correct.