Question

If $\alpha \ne \beta $ and ${\alpha ^2} = 5\alpha - 3{\text{ and }}{\beta ^2} = 5\beta - 3$ then the quadratic equation whose roots are $\dfrac{\alpha }{\beta }{\text{ and }}\dfrac{\beta }{\alpha }$ is:

Answer 1

Hint: Here as we are given that ${\alpha ^2} - 5\alpha + 3 = 0{\text{ and }}{\beta ^2} - 5\beta + 3 = 0$ and therefore we can say that $\alpha {\text{ and }}\beta $ are the roots of the quadratic equation ${x^2} - 5x + 3 = 0$.
Now we can find the sum of roots and product of the roots of the quadratic equation and substitute it in the equation ${x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$ and get the desired equation.

Complete Step by Step Solution:
Here we are given that ${\alpha ^2} = 5\alpha - 3{\text{ and }}{\beta ^2} = 5\beta - 3$
So we can also write this in the form ${\alpha ^2} - 5\alpha + 3 = 0{\text{ and }}{\beta ^2} - 5\beta + 3 = 0$
From this we can say that $\alpha {\text{ and }}\beta $ are the roots of the quadratic equation ${x^2} - 5x + 3 = 0$ because roots are just the values that can satisfy the given quadratic equation.
Hence we can say that as $\alpha {\text{ and }}\beta $ are the roots of the quadratic equation ${x^2} - 5x + 3 = 0$ so we can find the sum of roots and the product of roots.
Sum of roots$\left( {\alpha + \beta } \right)$$ = - \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}$$ = = - \dfrac{{ - 5}}{1} = 5$
Product of the roots$\left( {\alpha \beta } \right)$$ = \dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{x^2}}} = \dfrac{3}{1} = 3$
Now we want the roots as $\dfrac{\alpha }{\beta }{\text{ and }}\dfrac{\beta }{\alpha }$ and therefore we can find their sum and product and put in the general equation form which is:
${x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$
Sum of roots that are required$ = \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$
Now we can write ${\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta $
So we can write \[{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \]
So we can put these values in the sum of roots required:
Sum of roots that are required$ = \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$$ = \dfrac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }}$
Now we know the value of:
Sum of roots$\left( {\alpha + \beta } \right)$$ = - \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}$$ = = - \dfrac{{ - 5}}{1} = 5$
Product of the roots$\left( {\alpha \beta } \right)$$ = \dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{x^2}}} = \dfrac{3}{1} = 3$
Substituting it in above sum we will get:
Sum of roots that are required$ = \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$$ = \dfrac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }}$.
Sum of roots$ = \dfrac{{{{\left( { - 5} \right)}^2} - 2\left( 3 \right)}}{3} = \dfrac{{25 - 6}}{3} = \dfrac{{19}}{3}$.
Product of the roots which are given is $ = \dfrac{\alpha }{\beta } \times \dfrac{\beta }{\alpha } = 1$
Now we have got the sum as well as the product of the roots. Now we can find the quadratic equation by using the general property of the equation which is given as;
${x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$
$
  {x^2} - \dfrac{{19}}{3}x + 1 = 0 \\
  3{x^2} - 19x + 3 = 0 \\
 $

Hence we get the required quadratic equation as $3{x^2} - 19x + 3 = 0$.

Note:
Here the main point is in finding the roots’ sum as well as the product by using the roots’ sum and product that are given by the quadratic equation. Then we must know he general form of the quadratic equation which is ${x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$.