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Question

Answers

(a)8

(b)16

(c)32

(d)64

Answer
Verified

Hint: It is given that $\alpha ,\beta $ are the roots of the given quadratic equation so we can write sum and product of the roots by using the sum and product of the roots formula for the quadratic equation $a{{x}^{2}}+bx+c=0$ which is; sum of the roots equal to $-\dfrac{b}{a}$ and product of the roots equal to $\dfrac{c}{a}$. Now, we have to evaluate ${{\alpha }^{5}}+{{\beta }^{5}}$ which we can write as $\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$. Using sum and the product of the roots we can evaluate this expression.

__Complete step-by-step answer:__

The quadratic equation given in the above question is:

${{x}^{2}}-2x+4=0$

It is also given that $\alpha ,\beta $ are the roots of the above quadratic equation.

We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$ we can write sum and product of the roots as follows:

Sum of the roots $=-\dfrac{b}{a}$

Product of the roots $=\dfrac{c}{a}$

Using this formula, we can find the sum of the roots and product of the roots of the given equation.

$\begin{align}

& \alpha +\beta =2..........Eq.(1) \\

& \alpha \beta =4..........Eq.(2) \\

\end{align}$

We have to find the value of ${{\alpha }^{5}}+{{\beta }^{5}}$ which we can write as follows:

${{\alpha }^{5}}+{{\beta }^{5}}=\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$……… Eq. (3)

In the above equation, we can write:

${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $

Substituting the values from eq. (1) & eq. (2) in the above equation we get,

$\begin{align}

& {{\alpha }^{2}}+{{\beta }^{2}}={{\left( 2 \right)}^{2}}-2\left( 4 \right) \\

& \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=4-8=-4 \\

\end{align}$

\[\begin{align}

& {{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right) \\

& \Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta \right) \\

\end{align}\]

Substituting the values of ${{\alpha }^{2}}+{{\beta }^{2}}$ in the above equation we get,

\[{{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( -4-\alpha \beta \right)\]

Substituting the values from eq. (1) & eq. (2) in the above equation we get,

\[\begin{align}

& {{\alpha }^{3}}+{{\beta }^{3}}=\left( 2 \right)\left( -4-4 \right) \\

& \Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}=2\left( -8 \right)=-16 \\

\end{align}\]

From the above solution we have got the value of;

${{\alpha }^{2}}+{{\beta }^{2}}=-4$

\[{{\alpha }^{3}}+{{\beta }^{3}}=-16\]

Now, substituting the above values in eq. (3) we get,

$\begin{align}

& {{\alpha }^{5}}+{{\beta }^{5}}=\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right) \\

& \Rightarrow {{\alpha }^{5}}+{{\beta }^{5}}=\left( -16 \right)\left( -4 \right)-{{\left( 4 \right)}^{2}}\left( 2 \right) \\

& \Rightarrow {{\alpha }^{5}}+{{\beta }^{5}}=64-32=32 \\

\end{align}$

From the above solution, we have got the value of ${{\alpha }^{5}}+{{\beta }^{5}}$ as 32.

Hence, the correct option is (c).

Note: You might have thought why we have reduced ${{\alpha }^{5}}+{{\beta }^{5}}$ to $\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$. The reason behind is that in this reduced form, $\alpha ,\beta $ is written in the sum and product form. We know the value of $\alpha +\beta \And \alpha \beta $ so we can easily solve the expression $\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$ and hence get the value of ${{\alpha }^{5}}+{{\beta }^{5}}$. If we directly use quadratic formulas to find roots of the equation, we will get complex numbers as roots. So, it is advised not to go through this process.

The quadratic equation given in the above question is:

${{x}^{2}}-2x+4=0$

It is also given that $\alpha ,\beta $ are the roots of the above quadratic equation.

We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$ we can write sum and product of the roots as follows:

Sum of the roots $=-\dfrac{b}{a}$

Product of the roots $=\dfrac{c}{a}$

Using this formula, we can find the sum of the roots and product of the roots of the given equation.

$\begin{align}

& \alpha +\beta =2..........Eq.(1) \\

& \alpha \beta =4..........Eq.(2) \\

\end{align}$

We have to find the value of ${{\alpha }^{5}}+{{\beta }^{5}}$ which we can write as follows:

${{\alpha }^{5}}+{{\beta }^{5}}=\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$……… Eq. (3)

In the above equation, we can write:

${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $

Substituting the values from eq. (1) & eq. (2) in the above equation we get,

$\begin{align}

& {{\alpha }^{2}}+{{\beta }^{2}}={{\left( 2 \right)}^{2}}-2\left( 4 \right) \\

& \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=4-8=-4 \\

\end{align}$

\[\begin{align}

& {{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}} \right) \\

& \Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( {{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta \right) \\

\end{align}\]

Substituting the values of ${{\alpha }^{2}}+{{\beta }^{2}}$ in the above equation we get,

\[{{\alpha }^{3}}+{{\beta }^{3}}=\left( \alpha +\beta \right)\left( -4-\alpha \beta \right)\]

Substituting the values from eq. (1) & eq. (2) in the above equation we get,

\[\begin{align}

& {{\alpha }^{3}}+{{\beta }^{3}}=\left( 2 \right)\left( -4-4 \right) \\

& \Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}=2\left( -8 \right)=-16 \\

\end{align}\]

From the above solution we have got the value of;

${{\alpha }^{2}}+{{\beta }^{2}}=-4$

\[{{\alpha }^{3}}+{{\beta }^{3}}=-16\]

Now, substituting the above values in eq. (3) we get,

$\begin{align}

& {{\alpha }^{5}}+{{\beta }^{5}}=\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right) \\

& \Rightarrow {{\alpha }^{5}}+{{\beta }^{5}}=\left( -16 \right)\left( -4 \right)-{{\left( 4 \right)}^{2}}\left( 2 \right) \\

& \Rightarrow {{\alpha }^{5}}+{{\beta }^{5}}=64-32=32 \\

\end{align}$

From the above solution, we have got the value of ${{\alpha }^{5}}+{{\beta }^{5}}$ as 32.

Hence, the correct option is (c).

Note: You might have thought why we have reduced ${{\alpha }^{5}}+{{\beta }^{5}}$ to $\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$. The reason behind is that in this reduced form, $\alpha ,\beta $ is written in the sum and product form. We know the value of $\alpha +\beta \And \alpha \beta $ so we can easily solve the expression $\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)-{{\alpha }^{2}}{{\beta }^{2}}\left( \alpha +\beta \right)$ and hence get the value of ${{\alpha }^{5}}+{{\beta }^{5}}$. If we directly use quadratic formulas to find roots of the equation, we will get complex numbers as roots. So, it is advised not to go through this process.

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