Question

i want answer and direct copied answer from internet will be reported ..be aware​

Answer 1

Answer:

$\huge\underline\bold\red{Answer!!}$

$3x {}^{2} + kx + 81 = 0$

$let \: the \: roots \: be \: \alpha \: and \: \alpha {}^{2}$

$\alpha + \alpha {}^{2} = \frac{ - k}{3} ........................(1)$

$\alpha + \alpha {}^{2} = \frac{81}{3}$

$= > \: \: \: \: \: \: \: \: k {}^{3} = 27$

$= > \: \: \: \: \: \: \: \: k {}^{3} = 3.....................(2)$

$sub \: (2) \: in \: (1) \: we \: get$

$\: \: \: \: \: \: \: \: \: \: \: \: \: 3 + 3 {}^{2} = \frac{ - k}{3}$

$= > \: \: \: \: \: \: \: \: (3 + 9)3 = - k$

$= > \: \: \: \: \: \: \: \: \: \: \: \: \: \: k = - 36$

Hope it helps you ❤️ ✨

Answer 2

Your Answer:

Given:-

• Quadratic Equation = $\tt 3x^2 + kx + 81 = 0$.
• One zero is the square of another.

Your Answer:

We have

$\tt \alpha = \beta ^2$

Where alpha and beta are the zeros of Equation

We know that

$\tt \alpha \times \beta = c/a \\\\ \Rightarrow \tt \beta ^2 \beta = 81/3 \\\\ \tt \Rightarrow \beta ^3 = 27 \\\\ \tt \Rightarrow \beta = 3$

$\tt So, \alpha = 3^2 =9$

Now,

We also know that

$\tt \alpha + \beta = \dfrac{-b}{a}$

So,  replacing values

$\tt 9+3 = \dfrac{-k}{3} \\\\ \tt \Rightarrow 12 = \dfrac{-k}{3} \\\\ \tt \Rightarrow 12 \times 3 = -k \\\\ \tt \Rightarrow -36 = k$

So, value of k = -36