Question

If the difference of the zeroes of the quadratic polynomial is 5/2 and the difference of their

cubes is 65/8, find the quadratic polynomial.​

Answer 1

Answer:

$2 {x}^{2} - 3x - 2 = 0 \ \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{2} + 3x - 2 = 0$

Step-by-step explanation:

$\alpha - \beta = \frac{5}{2} \\ \\ { \alpha }^{3} - { \beta }^{3} = \frac{65}{8} \\ \\ ( \alpha - \beta )( { \alpha }^{2} + { \beta }^{2} + \alpha \beta ) = \frac{65}{8} \\ \\ \frac{5}{2} ( {( \alpha - \beta) }^{2} +3 ( \alpha \beta ) = \frac{65}{8} \\ \\ ( \frac{25}{4} +3 \alpha \beta ) = \frac{13}{4} \\ \\ \alpha \beta = - 1 \\ \\ ( { \alpha + \beta )}^{2} = {( \alpha - \beta) }^{2} + 4 \alpha \beta \\ \\ {( \alpha + \beta ) }^{2} = \frac{25}{4} + 4( - 1) \\ \\ {( \alpha + \beta )}^{2} = \frac{25 - 16}{4} = \frac{9}{4} \\ \\ \alpha + \beta = \frac{3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \alpha + \beta = \frac{ - 3}{2} \\ \\ required \: \: equation \: \: \: \: \: \: \: {x}^{2} - \frac{3}{2} x - 1 = 0 \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \: {x}^{2} + \frac{3}{2} x - 1 = 0 \\ \\ \\ 2 {x}^{2} - 3x - 2 = 0 \ \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{2} + 3x - 2 = 0$