Question

If $\alpha ,\beta$ and $\gamma$ are the roots of the equation $2x^3-3x^2+6x+1=0$, the ${\alpha }^2+{\beta }^2+{\gamma }^2$ is equal to

• A. $-\frac{15}{4}$
• B. $\frac{15}{4}$
• C. $\frac{9}{4}$
• D. $4$

Answer 1

The correct option is A

$-\frac{15}{4}$

Explanation for the correct option:

Step 1. Find the value of ${\alpha }^2+{\beta }^2+{\gamma }^2$:

Given, $\alpha ,\beta$ and $\gamma$ are the roots of the equation $2x^3-3x^2+6x+1=0$

As we know,

$S_1=$[coefficient of $x^2$] / [coefficient of $x^3$]

$$\begin{gathered} =\alpha +\beta +\gamma \\ =-\left(\frac{-3}{2}\right) \\ =\frac{3}{2} \end{gathered}$$

$S_2=$[coefficient of $x$] / [coefficient of $x^3$]

$$\begin{gathered} =\alpha \beta +\beta \gamma +\gamma \alpha \\ =-\frac{1}{2} \end{gathered}$$

$S_3=$[coefficient of constant term] / [coefficient of $x^3$]

$$\begin{gathered} =\alpha \beta \gamma \\ =-\frac{1}{2} \end{gathered}$$

As we know,

${\mathit{\alpha }}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{\gamma }}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\gamma }\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{2}\mathbf{}\mathbf{(}\mathit{\alpha }\mathit{\beta }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathit{\gamma }\mathbf{}\mathbf{+}\mathbf{}\mathit{\alpha }\mathit{\gamma }\mathbf{)}$

Step 2. Put the values of $S_1,S_2,S_3$ in above equation, we get

$\begin{array}{rcl}\therefore {\alpha }^2+{\beta }^2+{\gamma }^2& =& {(\alpha +\beta +\gamma )}^2–2(\alpha \beta +\beta \gamma +\alpha \gamma )\\ & =& {\left(\frac{3}{2}\right)}^2-2\times 3\\ & =& \frac{9}{4}-6\\ & =& \mathbf{-}\frac{\mathbf{15}}{\mathbf{4}}\end{array}$

Hence, Option ‘A’ is Correct.