Question

Can anyone solve this question PLZZZZZ. I will mark him/her as brainlist......​

Answer 1

$\tt \qquad ( {y}^{2} + 5y)( {y}^{2} + 5y - 2) - 24 = 0 \\ \\ \implies \tt {y}^{2} ( {y}^{2} + 5y - 2) + 5y( {y}^{2} + 5y - 2) - 24 = 0\\ \\ \implies \tt {y}^{4} + 5 {y}^{3} - 2 {y}^{2} + 5 {y}^{3} + 25 {y}^{2} - 10y - 24 = 0\\ \\ \implies \tt {y}^{4} + (5 + 5) {y}^{3} + (25 - 2) {y}^{2} - 10y - 24 = 0\\ \\ \tt \implies {y}^{4} + 10 {y}^{3} + 23 {y}^{2} - 10y - 24 = 0$
Therefore,

$\tt sum \: \: of \: \: zeroes. \: \alpha + \beta + \gamma + \delta = - \frac{10}{1} = ( - 10) \\ \\ \tt sum \: \: of \: \: zeroes \: \: taken \: \: product \: \: of \: \: two \: \: zeroes \: \: at \: \: a \: \: time. \\ \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{23}{1} = 23 \\ \\ \\ \tt sum \: \: of \: \: zeroes \: taken \: \: product \: \: of \: \: three \: \: at \: \: a \: \: time. \\ \alpha \beta \gamma + \alpha \beta \delta + \beta \gamma \delta + \alpha \gamma \delta = - \frac{( - 10)}{1} = 10 \\ \\ \\ \tt product \: \: of \: \: zeroes. \: \alpha \beta \gamma \delta = - \frac{( - 24)}{1} = 24$