Answer
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Hint: To attempt this question the knowledge of the concept of quadratic equation is must and also remember to use algebraic formula like ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$ and ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$ to simplify the equation, using this information you can approach the solution.
Complete step-by-step solution:
Now it is been given that $\alpha + \beta = 3$ and ${\alpha ^3} + {\beta ^3} = 7$. We need to show that $\alpha $ and $\beta $ are the roots of $9{x^2} - 27x + 20 = 0$
Now ${\alpha ^3} + {\beta ^3} = 7$ so using ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
We can say that ${\alpha ^3} + {\beta ^3} = \left( {\alpha + \beta } \right)\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
Putting values from above we get
$7 = 3\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
$\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right) = \dfrac{7}{3}$
Now $\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
Since we know that ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$
So, we can rewrite the above equation as ${\left( {\alpha + \beta } \right)^2} - 3\alpha \beta $
${\left( {\alpha + \beta } \right)^2} - 3\alpha \beta = \dfrac{7}{3}$
Again, substituting the values, we get
$9 - \dfrac{7}{3} = 3\alpha \beta $
Hence $\alpha \beta = \dfrac{{20}}{9}$
Now if sum of roots and product of roots is given that the quadratic equation can be written as ${x^2} - (sum{\text{ of roots)x + product = 0}}$
Thus, the quadratic equation having roots $\alpha $ and $\beta $ is ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$
So, putting values we get
${x^2} - 3x + \dfrac{{20}}{9} = 0$
Therefore, on solving we get
$9{x^2} - 27x + 20 = 0$ which is the desired equation.
Note: Whenever we face such problems the key concept that needs to be in our mind is if somehow, we get the sum and the product of the roots then we can easily get the required quadratic equation. Hence simply accordingly to obtain sum and product of roots.
Complete step-by-step solution:
Now it is been given that $\alpha + \beta = 3$ and ${\alpha ^3} + {\beta ^3} = 7$. We need to show that $\alpha $ and $\beta $ are the roots of $9{x^2} - 27x + 20 = 0$
Now ${\alpha ^3} + {\beta ^3} = 7$ so using ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
We can say that ${\alpha ^3} + {\beta ^3} = \left( {\alpha + \beta } \right)\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
Putting values from above we get
$7 = 3\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
$\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right) = \dfrac{7}{3}$
Now $\left( {{\alpha ^2} + {\beta ^2} - \alpha \beta } \right)$
Since we know that ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$
So, we can rewrite the above equation as ${\left( {\alpha + \beta } \right)^2} - 3\alpha \beta $
${\left( {\alpha + \beta } \right)^2} - 3\alpha \beta = \dfrac{7}{3}$
Again, substituting the values, we get
$9 - \dfrac{7}{3} = 3\alpha \beta $
Hence $\alpha \beta = \dfrac{{20}}{9}$
Now if sum of roots and product of roots is given that the quadratic equation can be written as ${x^2} - (sum{\text{ of roots)x + product = 0}}$
Thus, the quadratic equation having roots $\alpha $ and $\beta $ is ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$
So, putting values we get
${x^2} - 3x + \dfrac{{20}}{9} = 0$
Therefore, on solving we get
$9{x^2} - 27x + 20 = 0$ which is the desired equation.
Note: Whenever we face such problems the key concept that needs to be in our mind is if somehow, we get the sum and the product of the roots then we can easily get the required quadratic equation. Hence simply accordingly to obtain sum and product of roots.
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