Question

Very much difficult and challenger question for you
You means only for maths genius

The answer is 2/3
But need with explination

Answer 1

$Answer \: \: \\ \\ Given \: Quadratic \: polynomial \: \: is \\ \\ f(t) = kt {}^{2} + 2t + 3k \\ \\ the \: given \: polynomial \: have \: two \: roots \\ let \: those \: roots \: be \: \: \: \: \alpha \: \: \: \: and \: \: \: \: \: \beta \\ \\ \alpha + \beta = \frac{ - 2}{k} \: \: \: \: \: \: and \: \: \: \: \: \alpha \beta = \frac{3k}{k} \\ \\ ACCORDING \: TO \: \: the\: \: given \: question \: \\ \\ \alpha + \beta = \alpha \beta \\ \\ \frac{ - 2}{k} = \frac{3k}{k} \\ \\ \frac{ - 2}{k} = 3 \\ \\ k = \frac{ - 2}{3} \\ \\ NOTE \: \\ \\ for \: a \: general \: Quadratic \: polynomial \: say \\ \\ f(x) = ax {}^{2} + bx + c \: \\ \\ \alpha + \beta = \frac{ - b}{a} \: \: \: \: and \: \: \: \: \: \alpha \beta = \frac{c}{a} \\ \\ where \: \: \: \alpha \: \: and \: \: \beta \: are \: its \: zeroes \:$

Answer 2

$\huge\textbf\red{Answer:-}$

$\huge\textbf{Quadric-Eqation:}$

$\huge\textbf{Here,}$

Ft is equal to kt²+2t+3k this is equal to k

We need to find the value of K,here

$(F-t)=( kt2 +2t +3t)$

$\huge\textbf{Expalnations:}$

$\huge\textbf{Considering,}$

$\Large\text{(a=k)}$

$\Large\text{(b=2)}$

$\Large\text{(c=3k)}$

$Zeros= \frac{ - b}{c}$

Now, add Alpha and Beta (α+β)$Alpha +Beta= \frac{ - 2}{k}$

$(Zeros= \frac{c}{a}) \\ </p><p>(alpha=beta)=( \frac{3k}{k} )$

α and β[Alpha add Beta=Alpha Beta(αβ) ]$(Alpha+Beta=Alpha=Beta)$

$(3= \frac{ - 2}{kk}) \\ </p><p>(k= \frac{ - 2}{3} )$

α=Alpha

β=Beta

$Hence, Value \: of (K= \frac{ - 2}{3})$

$\huge\textbf\red{Hope-It-Helps!!! }$