5. If [tex]$\alpha$[/tex] and [tex]$\beta$[/tex] are roots of the quadratic equation [tex]$5 x^2-3 x-1=0$[/tex], find a quadratic equation with integral coefficients which have roots:
(i) [tex]$\frac{1}{\alpha^2}$[/tex] and [tex]$\frac{1}{\beta^2}$[/tex]
(ii) [tex]$\frac{\alpha^2}{\beta}$[/tex] and [tex]$\frac{\beta^2}{\alpha}$[/tex]
Answer (2)
Find the sum and product of the roots α and β using Vieta's formulas: α + β = 5 3 and α β = − 5 1 .
For part (i), calculate the sum and product of the new roots α 2 1 and β 2 1 : α 2 1 + β 2 1 = 19 and α 2 1 ⋅ β 2 1 = 25 . The quadratic equation is x 2 − 19 x + 25 = 0 .
For part (ii), calculate the sum and product of the new roots β α 2 and α β 2 : β α 2 + α β 2 = − 25 72 and β α 2 ⋅ α β 2 = − 5 1 . The quadratic equation is 25 x 2 + 72 x − 5 = 0 .
The quadratic equations are x 2 − 19 x + 25 = 0 and 25 x 2 + 72 x − 5 = 0 .
Explanation
Problem Analysis We are given that α and β are roots of the quadratic equation 5 x 2 − 3 x − 1 = 0 . We need to find quadratic equations with integer coefficients that have roots (i) α 2 1 and β 2 1 and (ii) β α 2 and α β 2 .
Finding Sum and Product of Roots (i) We want to find a quadratic equation with roots α 2 1 and β 2 1 . Since α and β are roots of 5 x 2 − 3 x − 1 = 0 , we can use Vieta's formulas to find the sum and product of the roots. The sum of the roots is α + β = 5 − ( − 3 ) = 5 3 , and the product of the roots is α β = 5 − 1 = − 5 1 .
Sum of New Roots Now, let's find the sum and product of the new roots α 2 1 and β 2 1 . The sum is α 2 1 + β 2 1 = α 2 β 2 α 2 + β 2 = ( α β ) 2 ( α + β ) 2 − 2 α β . Substituting the values of α + β and α β , we get ( − 5 1 ) 2 ( 5 3 ) 2 − 2 ( − 5 1 ) = 25 1 25 9 + 5 2 = 25 1 25 9 + 10 = 25 1 25 19 = 19 .
Quadratic Equation for Part (i) The product of the new roots is α 2 1 ⋅ β 2 1 = ( α β ) 2 1 = ( − 5 1 ) 2 1 = 25 1 1 = 25 . Therefore, the quadratic equation with roots α 2 1 and β 2 1 is x 2 − ( s u m o f roo t s ) x + ( p ro d u c t o f roo t s ) = 0 , which is x 2 − 19 x + 25 = 0 .
Sum of New Roots (ii) Now, we want to find a quadratic equation with roots β α 2 and α β 2 . Let's find the sum and product of these roots. The sum is β α 2 + α β 2 = α β α 3 + β 3 = α β ( α + β ) ( α 2 − α β + β 2 ) = α β ( α + β ) (( α + β ) 2 − 3 α β ) . Substituting the values of α + β and α β , we get − 5 1 ( 5 3 ) (( 5 3 ) 2 − 3 ( − 5 1 )) = − 5 1 ( 5 3 ) ( 25 9 + 5 3 ) = − 5 1 ( 5 3 ) ( 25 9 + 15 ) = − 5 1 ( 5 3 ) ( 25 24 ) = − 5 ( 5 3 ) ( 25 24 ) = − 3 ( 25 24 ) = − 25 72 .
Quadratic Equation for Part (ii) The product of the new roots is β α 2 ⋅ α β 2 = α β = − 5 1 . Therefore, the quadratic equation with roots β α 2 and α β 2 is x 2 − ( s u m o f roo t s ) x + ( p ro d u c t o f roo t s ) = 0 , which is x 2 − ( − 25 72 ) x − 5 1 = 0 . Multiplying by 25 to get integer coefficients, we have 25 x 2 + 72 x − 5 = 0 .
Final Answer Thus, the quadratic equation with roots α 2 1 and β 2 1 is x 2 − 19 x + 25 = 0 , and the quadratic equation with roots β α 2 and α β 2 is 25 x 2 + 72 x − 5 = 0 .
Examples
Understanding the relationship between the roots and coefficients of a quadratic equation is useful in various fields, such as physics and engineering. For example, in circuit analysis, the roots of a characteristic equation determine the stability of a circuit. If you know the roots, you can determine the equation, and vice versa. This allows engineers to design circuits with specific behaviors. Similarly, in control systems, understanding the roots of the system's transfer function helps in designing controllers that ensure the system is stable and performs as desired. The ability to manipulate and transform these roots, as demonstrated in this problem, provides a powerful tool for analyzing and designing such systems.
To find the quadratic equations with specific roots, we use the properties of the original quadratic equation's roots. The equations are x 2 − 19 x + 25 = 0 for roots α 2 1 and β 2 1 , and 25 x 2 + 72 x − 5 = 0 for roots β α 2 and α β 2 .
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