Question

Find the sum of the roots of equation
2x^3-5x^2+ 6x - 123 = 0.​

Answer 1

Answer:

$\frac{5}{2}$

Step-by-step explanation:

$2 {x}^{3} - 5 {x}^{2} + 6x - 123 \\ \\ \alpha + \beta + \gamma = \frac{ - b}{a} \\ \alpha + \beta + \gamma = \frac{ - ( - 5)}{2} \\ \\ \\ \alpha + \beta + \gamma = \frac{5}{2}$

Answer 2

Step-by-step explanation:

$\huge\underline{\underline{\sf ANSWER}} \\ let.... \\ \alpha , \beta \: and \gamma \: be \: roots.... \\ \\ sum \: of \: roots = \frac{ - b}{a} \\ \alpha + \beta + \gamma = -(\frac{-5}{2}) \\ \alpha + \beta + \gamma = \frac{5}{2} \\ \\ sum \: of \: product \: of \: 2 \: roots = \frac{c}{a} \\ \alpha \beta + \alpha \gamma + \beta \gamma = \frac{6}{2} = 3 \\ \\ product \: of \: roots = \frac{ - d}{a} \\ \alpha \beta \gamma = -( \frac{-123}{2}) \\ \alpha \beta \gamma = \frac{123}{2}$

$if \: this \: helps \: don't \: forget \: \\ to \: thank \: this \: answer$