Question

(iii) If a and ß are the roots of the quadratic equation 3x2 + kr + 8.=0 and alpha:beta=2:3 , then find the k value.​

Answer 1

HEY MATE HERE'S THE ANSWER........ ❤️❤️❤️❤️❤️

GIVEN :

$3x {}^{2} + kx + 8 = 0 \\ \\ \alpha \: \: and \: \: \beta \: \: are \: \: its \: \: root \: \\ \\ \ \frac{ \alpha }{ \beta } = \frac{2}{3}$

TO FIND : VALUE OF k

SOLUTION :

$we \: \: know \: \: \\ product \: \: of \: \: zeroes \: = \alpha \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ c}{a} \\ \\ therefore \: ...... \\ \\ \alpha \beta = \frac{ 8}{3} ...........eq {}^{1} \\ also \\ \\ \frac{ \alpha }{ \beta } = \frac{2}{3} ..........(given) \\ \\ now \: multiply \: given \: and \: eq {}^{1} \\ \\ you \: will \: get \: \\ \alpha {}^{2} = \frac{16}{9} \\ taking \: square \: root \: both \: sides \\ \\ \alpha = + \frac{4}{3} \: \: or \: - \frac{4}{3} \\ \\ \: \: you \: will \: get \: two \: cases \: \\ \\ \\ case \: 1 \\ \\ when \: \alpha = \frac{ - 4}{3} \\ \\ you \: will \: get \: \beta = - 2 \\ \\ case2 \\ \\ when \: \alpha = + \frac{4}{3} \\ you \: will \: get \: \beta = 2$

Now,,

K can be both positive and negative.

$using \: \\ sum \: of \: zeroes \: = \alpha + \beta = \frac{ - b}{a} \\ \\ case \: 1 \\ \frac{-k}{3} \: = \frac{ - 4}{3} + ( - 2) \\ = > \frac{ - 10}{3 } \\ \\ therefore \: k = 10 \\ \\ similarly \: \\ k = - 10$

Hence k can be +10 or - 10

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Answer 2

Answer:

α/β = 2/3 --> ( i )

[ We use the "Co-efficient - Zeroes relation / also known as Viete's Relation ] --> 

--> α * β = 8/3 --> ( ii )

--> Multiplying ( i ) with ( ii ) --> 

 ---> α² = 16 / 9

  => α = ± ( 4/3 )

Correspondingly, β = ± 2

Further, we have the relation, α + β = -k / 3

         => ± [ 4/3 + 2 ] = -k / 3

         => ± [ 10 / 3 ] = - k / 3

         => k = - 10 or +10 

However, since, k > 0, k = 10 for α = -4/3 || β = -2 is considered the reqd.