Question

If the roots of the quadratic equation $x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$ are $\alpha$ and $\beta$, then the value of ${\alpha }^{96}\left({\alpha }^{12}-1\right)+{\beta }^{96}\left({\beta }^{12}-1\right)$

• A. $50.3^{24}$
• B. $51.3^{24}$
• C. $52.3^{24}$
• D. $104.3^{24}$

Answer 1

The correct option is C

$52.3^{24}$

Step 1: Solve for the values of $\alpha$ and $\beta$

Given that the roots of the quadratic equation $x^2+3^{\frac{1}{4}}x+3^{\frac{1}{2}}=0$ are $\alpha$ and $\beta$

$\therefore x^2+3^{\frac{1}{2}}=-3^{\frac{1}{4}}x$

Squaring on both sides, we get,

$\begin{array}{crcl}& {\left(x^2+3^{\frac{1}{2}}\right)}^2& =& {\left(-3^{\frac{1}{4}}x\right)}^2\\ \Rightarrow & x^4+2.x^2.3^{\frac{1}{2}}+3& =& 3^{\frac{1}{2}}{x}^2\\ \Rightarrow & x^4+3& =& -3^{\frac{1}{2}}{x}^2\end{array}$

Squaring on both sides, we get,

$\begin{array}{crcl}& {\left(x^4+3\right)}^2& =& {\left(-3^{\frac{1}{2}}{x}^2\right)}^2\\ \Rightarrow & x^8+2.x^4.3+9& =& 3x^4\\ \Rightarrow & x^8& =& -9-3x^4\end{array}$

$\because \alpha ,\beta$ are the roots of the equation, we can write as

${\alpha }^8=-9-3{\alpha }^4$ and ${\beta }^8=-9-3{\beta }^4$

Step 2: Solve for the required value

Consider ${\alpha }^8=-9-3{\alpha }^4$

Multiply with ${\alpha }^4$on both sides

$\begin{array}{crclc}& {\alpha }^4\left({\alpha }^8\right)& =& {\alpha }^4\left(-9-3{\alpha }^4\right)& \\ \Rightarrow & {\alpha }^{12}& =& -9{\alpha }^4-3{\alpha }^8& \\ \Rightarrow & {\alpha }^{12}& =& -9{\alpha }^4-3\left(-9-3{\alpha }^4\right)& \left[\because {\alpha }^8=\left(-9-3{\alpha }^4\right)\right]\\ \Rightarrow & {\alpha }^{12}& =& -9{\alpha }^4+27+9{\alpha }^4& \\ \Rightarrow & {\alpha }^{12}& =& 27& \end{array}$

Similarly ${\beta }^{12}=27$

$\begin{array}{cccl}\therefore & {\alpha }^{96}\left({\alpha }^{12}-1\right)+{\beta }^{96}\left({\beta }^{12}-1\right)& =& {\left({\alpha }^{12}\right)}^8\left({\alpha }^{12}-1\right)+{\left({\beta }^{12}\right)}^8\left({\beta }^{12}-1\right)\\ & & =& {\left(27\right)}^8\left(27-1\right)+{\left(27\right)}^8\left(27-1\right)\\ & & =& {\left(3^3\right)}^8\left(26\right)+{\left(3^3\right)}^8\left(26\right)\\ & & =& \left(26+26\right).3^{24}\\ \Rightarrow & {\alpha }^{96}\left({\alpha }^{12}-1\right)+{\beta }^{96}\left({\beta }^{12}-1\right)& =& 52.3^{24}\end{array}$

Hence, the correct option is option(C) i.e. $52.3^{24}$.