Question

if the sum of zeroes of f(x)=2x3-3kæ2+ 4x-5
is 6 then find k ?​

Answer 1

Given:

(with correction)

f(x) = 2x³ - 3kx² + 4x - 5

Sum of zeroes = 6

To find:

the value of k

Answer:

Sum of zeroes = - (coefficient of x²) / (coefficientof x³)

= - b/a

$let \: zeroes \: be \: \alpha \: and \: \beta$

So,

$\alpha + \beta = \frac{ - b}{a}$

$\alpha + \beta = \frac{ - ( - 3k)}{2} \\ = \frac{3k}{2}$

also,

$\alpha + \beta = 6 \: \: ({from \: given})$

Now,

=> 6 = 3k/2

=> 12=3k

=> k = 4

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Answer 2

Given:-

α+β=6

To Find:-

The value of K?

$\rule{200}{1}$

Answer:-

a=2

b=-3k

c=4

→α+β=$\frac{-b}{a}$

•Putting the Values•

→α+β=$\frac{-3k}{2}$

[°•°α+β=6]

→6=$\frac{-(-3k)}{2}$

♦Cross Multiply LHS&RHS♦

→6×2=3k

→12=3k

→$\frac{12}{3}$=k

→4=k

★The value of k=4★

$\rule{200}{8}$