Question

Genius type question
solve it maths lovers

Answer 1

$\huge\star\underline{\mathcal\color{brown}{HELLO\:MATE}}\star$

$\color{brown}{HERE\:IS\:YR\:ANS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\underline\color{Green}{SOL}$

SEE THE GIVEN ATTACHMENT

IT IS SO SIMPLE TO SOLVE

$\fbox\color{brown}{HOPE\:IT\:HELPS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\huge{\mathcal{THANKS}}$

$<marquee >☝️ABHI♥️☝️$

Answer 2

$\huge{\red{ANSWER}}$

$Answer \: \\ \\ Given \: cubic\: \: Polynomial \: is \\ \\ p(x) = x {}^{3} - 5x {}^{2} - 2x + 24 \\ this \: equation \: contains \: three \: roots \: \\ let \: those \: roots \: be \: \: \alpha \: \: \: \: \beta \: \: \: \: and \: \: \: \gamma \\ \\ \alpha + \beta + \gamma = \frac{5}{1} \: \: \: \: \: \alpha \beta \gamma = \frac{ - 12 }{1} \\ \\ and \: \: \: \: \: \: \alpha \beta + \beta \gamma + \alpha \gamma = \frac{ - 2}{1} \\ \\ According \: \: to \: the \: question \: \: \: \: \alpha \beta = 12 \\ \\ \alpha + \beta + \gamma = 5 \: \: \: ... \: \: Equation \: \: \: i \\ \\ \alpha \beta \gamma = - 24 \: \: ...Equation \: \: \: ii \\ \\ \alpha \beta + \beta \gamma + \alpha \gamma = - 2\: \: \: ...Equation \: \: \: iii \\ \\ put \: \: value \: \: of \: \: \: \alpha \beta = 12 \: \: \: in \: \: Equation \: ii \: we \: have \\ \\ 12 \times \gamma = - 24 \\ \\ \gamma = \frac{ - 24}{12} \\ \\ \gamma = - 2 \\ \\ put \: value \: \: of \: \: \gamma \: \: in \: Equation \: i \: we \: have \\ \\ \alpha + \beta - 2 = 5 \\ \\ \alpha + \beta = 7 \: \: \: ....Equation \: \: \: iv \\ \\ \beta = 7 - \alpha \\ \\ now \: put \: the \: value \: of \: \gamma \: \:and \: \: \beta \: in \: equation \: iii \: we \: have \\ \\ \alpha (7 - \alpha ) - 2(7 - \alpha ) - 2 \alpha = - 2 \\ \\ 7 \alpha - \alpha {}^{2} - 14 + 2\alpha - 2 \alpha = - 2 \\ \\ - \alpha {}^{2} + 7 \alpha - 12 = 0 \\ \\ \alpha {}^{2} - 7 \alpha + 12 = 0 \\ \\ \alpha {}^{2} - 4 \alpha - 3 \alpha + 12 = 0 \\ \\ \alpha ( \alpha - 4) - 3( \alpha - 4) = 0 \\ \\ \alpha = 4 \: \: \: \: \: \: or \: \: \: \: \alpha = 3 \\ \\ for \: \: \: \: \: \alpha = 4 \: \: \: \: \: \: \: \: \beta = 3 \\ \\ for \: \: \: \: \alpha = 3 \: \: \: \: \: \: \beta = 4 \\ \\ for \: \: \: \alpha = 3 \: \: \: and \: \: \: \beta = 4 \: \: \: \: \gamma = - 2 \\ \\ for \: \: \: \: \alpha = 4 \: \: \: \: and \: \: \: \: \beta = 3 \: \: \: \: \gamma \: = - 2 \\ \\ \\ therefore \: \: zeroes \: \: of \: \: this \: cubic \: polynomial \: \\ are \: \: \: \: \: 4 \: \: \: \: 3 \: \: \: \: and \: \: \: - 2 \\ \\ or \\ \\ 3 \: \: \: 4 \: \: \: \: and \: \: \: \: - 2 \\ \\ NOTE \: \\ \\ for \: a \: general \: cubic \: polynomial \: say \\ \\ f(x) = ax {}^{3} + bx {}^{2} + cx + d \\ \\ \alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ \alpha \beta \gamma = \frac{ - d}{a} \\ \\ \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \\ where \: \: \alpha \: \: \: \: \: \beta \: \: \: and \: \: \gamma \: \: are \: its \: zeroes \: .$